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# Homotopy and homotopy equivalence

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## 1 Deforming spaces vs. deforming maps

What do the objects below have in common?

The answer we have been giving is: they all have one hole. However, there is a profound reason why they must all have one hole. These space are homeomorphic!

The reasoning, still not fully justified, is transparent: $$X\approx Y \Rightarrow H(X)\cong H(Y).$$

Now, let's choose a broader collection. This time the spaces aren't homeomorphic, but do they have anything in common?

The answer again is: they all have one hole. But, again, maybe there is a profound reason why they all have one hole.

Is there a relation between two topological spaces, besides homeomorphism, that ensures that they would have the same count of topological features? There is indeed an equivalence relation that produces the same result for a much broader class of spaces: $$X\sim Y \Rightarrow H(X)\cong H(Y).$$

Informally, we say that one space can be “deformed into” the other.

Let's try to understand the actual mathematics behind these words, with the help of this juxtaposition:

the cylinder $X={\bf S}^1 \times {\bf I}$ vs. the circle $Y={\bf S}^1$.

The first meaning of the word “deformation” is a transformation that is gradual. Unlike a homeomorphism, which is instantaneous, this transformation is stretched over time through a continuum of intermediate states:

Let's take a look at the natural maps between these two spaces. The first is the projection $p:X\to Y$ of the cylinder along its axis and the second is the embedding $e:Y\to X$ of the circle into the cylinder as one of its boundary circles:

Both preserve the hole even though neither one is a homeomorphism.

Let's assume that $X$ is the unit cylinder in ${\bf R}^3$ and $Y$ is the unit circle in ${\bf R}^2 \subset {\bf R}^3$. Then the formulas are simple: $$p(x,y,z)= (x,y),\ e(x,y)=(x,y,0).$$ Let's consider the compositions of these maps:

• $pe:Y\to Y$ is the identity $\operatorname{Id}_Y$;
• $ep:X\to X$ is the collapse, or self-projection, of the cylinder on its bottom edge.

The latter, even though not the identity, is related to $\operatorname{Id}_X$. This relation is seen through the continuum of maps that connects the two maps; we choose map $h_t:X\to X$ to shrink the cylinder -- within itself -- to height $t\in [0,1]$. The images of these maps are shown:

This is our main idea: we should interpret deformations of spaces via deformations of maps.

The formulas for these maps are simple: $$h_t(x,y,z)=(x,y,tz),\ t\in [0,1].$$ And, it is easy to confirm that we have what we need: $$h_0=ep,\ h_1=\operatorname{Id}_X.$$ Looking closer at these maps, we realize that

• $h_t$ is continuous for each $t$, but also
• $h_t$ are continuous, as a whole, over $t$.

Therefore, the transformation can be captured by a single map $$H(t,x,y,z)=h_t(x,y,z),$$ continuous with respect to the product topology.

The precise interpretation of this analysis is given by the two definitions below.

Definition. Two maps $f_0, f_1: X \to Y$ are called homotopic if there is a map $$F: [0,1]\times X \to Y$$ such that $$F(0,x) = f_0(x),\ F(1,x) = f_1(x),$$ for all $x\in X$. Map $F$ is called a homotopy between $f_0$ and $f_1$. The relation is denoted by: $$F:f_0 \simeq f_1,$$ or simply: $$f_0 \simeq f_1 .$$

Definition. Suppose that $X$ and $Y$ are topological spaces and $f: X \to Y,\ g: X \to Y$ are maps, and $fg$ and $gf$ are homotopic to the identity maps on $Y$ and $X$ respectively: $$fg \simeq \operatorname{Id}_{Y},\ gf \simeq \operatorname{Id}_{X},$$ then $f$ is called a homotopy equivalence. In this case, $X$ and $Y$ are called homotopy equivalent.

We set the latter aside for now and study properties of homotopy.

## 2 Homotopy

It is often hard to visualize a homotopy via its graph, unless the dimensions of the spaces are low. For example, below we have the graph of a homotopy (blue) between a constant map (orange) and the identity (purple) of an interval: $$c,\operatorname{Id}:[a,b]\to [a,b].$$

The diagram on the left demonstrates that $c \simeq \operatorname{Id}$ by showing the homotopy as a surface that “connects” these two maps.

On the right we also provide a more common way to illustrate a homotopy -- by plotting the “intermediate” maps. Those are, of course, simply the vertical cross-sections of this surface. The homotopy above is piece-wise linear and the one below is differentiable:

Theorem. Homotopy is an equivalence relation. For two given topological spaces $X$ and $Y$, the space $C(X,Y)$ of maps from $X$ to $Y$ is partitioned into equivalence classes: $$[f]:=\{g:X\to Y,\ g \simeq f\}.$$

Proof. 1. Reflexivity: do nothing. For $F:f\simeq f$, choose $H(t,x)=f(x)$.

2. Symmetry: reverse time. Given $H:f\simeq g$, choose $F(t,x)=H(1-t,x)$ for $F:g\simeq f$.

3. Transitivity: ? One needs to figure out this: $$F:f \simeq g,\ G: g\simeq h \Rightarrow H=?: f \simeq h.$$ We carry out these two processes consecutively but, since it has to be within the same time frame, twice as fast. We define: $$H(t,x):=\begin{cases} F(2t,x) & \text{ if } 0\le t\le 1/2 \\ G(2t-1,x) & \text{ if } 1/2 \le t\le 1.\\ \end{cases}$$ $\blacksquare$

Referring to the proof, these three new homotopies are called respectively:

• 1. a constant homotopy,
• 2. the inverse of a homotopy, and
• 3. the concatenation of two homotopies.

They are illustrated below:

Exercise. Provide the missing details of the proof.

Notation. We will use this notation for this quotient set: $$[X,Y]:=C(X,Y) / _{\simeq}.$$

Sometimes things are simple. Any function $f:X \to {\bf R}$ can be “transformed” into any other. In fact, a simpler idea is to push the graph of a given function $f$ to the $x$-axis:

We simply put: $$f_t(x):=tf(x).$$ Then, we use the fact that an equivalence relation creates a partition into equivalence classes, to conclude that $$[{\bf R},{\bf R}] = \{[0]\}.$$ In other words, all maps are homotopic.

We can still have an explicit formula for a homotopy between two functions $f,g$: $$F(t,x) = (1-t)f(x) + tg(x).$$

Exercise. Prove the continuity of $F$.

The same exact formula describes how one “gradually” slides $f(x):X\to Y$ toward $g(x)$ if $X$ is any topological space and $Y$ is a convex subset of ${\bf R}^n$, which guarantees that all convex combinations make sense:

This is called the straight-line homotopy.

A more general setting for this construction is the following. A vector space $V$ over ${\bf R}$ is called a topological vector space if it is equipped with a topology with respect to which its vector operations are continuous:

• addition: $V \times V \to V$, and
• scalar multiplication: ${\bf R} \times V \to V$.

Exercise. Prove that these are topological vector spaces: ${\bf R}^n$, $C[a,b]$. Hint: don't forget about the product topology.

Proposition. If $Y$ is a convex subset of a topological vector space, then all maps to $Y$ are homotopic: $\#[X,Y]=1.$

Exercise. Prove the proposition. Hint:

What if $Y$ is the disjoint union of $m$ convex sets in ${\bf R}^n$? Will we have: $$\#[X,Y]=m?$$ Yes, but only if $X$ is path-connected!

Exercise. Prove this statement and demonstrate that it fails without the path-connectedness requirement.

Exercise. Prove that if $Y$ isn't path-connected then $\#[X,Y] > 1$. Hint: consider constant maps:

Exercise. What if $X$ has $k$ path-components?

Next, what can we say about the homotopy classes of maps from the circle to the ring, or another circle?

The circles can be moved and stretched as if they were rubber bands:

Later we will classify these maps according to their homotopy classes.

Sometimes the homotopy is easy to specify. For example, one can gradually stretch this circle:

To get from $f_0$ (the smaller circle) to $f_0$ (the large circle), one goes through intermediate steps -- circles indexed by values between $0$ and $1$: $$f_0, f_{1/4}, f_{1/2}, f_{3/4}, f_1.$$ Generally: $$f_t(x) := (1+t)f_0(x).$$ It is clear that the right-hand side continuously depends on both $t$ and $x$.

Exercise. Suppose that two maps $f,g:X\to {\bf S}^1$ never take antipodal to each other points: $f(x)\ne -g(x),\ \forall x\in X$. Prove that $f,g$ are homotopic.

To summarize,

homotopy is a continuous transformation of a continuous function.

Homotopies help us tame the monster of a space filling curve.

Theorem. Every map $f:[0,1]\to S$, where $S$ is a surface, is homotopic to a map that isn't onto.

Proof. Suppose $Q$ is the point on $S$ that we want to avoid and $f(0)\ne Q,\ f(1)\ne Q$. Let $D$ be a small disk neighborhood around $Q$. Then $f^{-1}(D)\subset (0,1)$ is open, and, therefore, is the disjoint union of open intervals. Pick one of them, $(a,b)$. Then we have: $$\begin{array}{lll} A:=f(a) &\in \partial D,\\ B:=f(b) &\in \partial D,\\ f\big( (a,b) \big) &\subset D. \end{array}$$

Choose $$C \in \partial D \setminus \{A,B\}.$$ Now, $f([a,b])$ is compact and, therefore, closed. Then there is a neighborhood of $C$ disjoint with the path $f([a,b])$. We construct a homotopy that pushes this path away from $C$ towards the opposite side of $\partial D$. If we push far enough, the point $Q\in D$ won't lie on the path anymore.

This homotopy is “relative”: all the changes to $f$ are limited to the values $x\in (a,b)$. This allows us to build such a homotopy independently for every interval $(a,b)$ in the decomposition of $f^{-1}(D)$. Next, the fact that the condition $f(a)=A,\ f(b)=B$ is preserved under this homotopy allows us to stitch these “local” homotopies together one by one. The resulting homotopy $F$ pushes every piece of the path close to $Q$ away from $Q$.

This construction only works when there are finitely many such open intervals.

$\blacksquare$

Exercise. Show that the proof, as written, fails when we attempt to stitch together infinitely many “local” homotopies to build $F:[0,1]\times (0,1)\to S$. Hint: consider,

• $f^{-1}(D)=\cup_n (a_n,b_n),\ f(a_n)=A_n,\ f(b_n)=B_n,$
• $A_n\to A,\ B_n\to B=A,\ F(1,t_n)=-A,\ t_n\in (a_n,b_n)$.

Exercise. Fix the proof. Hint: choose only the intervals on which the path actually passes through $Q$.

Exercise. Provide an explicit formula for a “local” homotopy.

Exercise. Generalize the theorem as much as you can.

## 3 Connectedness

Comparing the familiar illustrations of path-connectedness on the left and a homotopy between two constant maps reveals that the former is a special case of the latter:

Indeed, suppose in $Y$ there is a path between two points $a,b$ as a function $p:[0,1] \to Y$ with $p(0)=a,p(1)=b$. Then the function $H: [0,1] \times X$ given by $H(t,x)=p(t)$ is a homotopy between these two constant maps.

Exercise. Sketch the graph of a homotopy between two constant maps defined on $[a,b]$.

To summarize, in a path-connected space all constant maps are homotopic.

Let's carry out a similar analysis for simple-connectedness. This condition was defined informally as every closed path can be deformed to a point.

By now we know what “deformed” means. Let's translate: $$\begin{array}{llll} \text{ Informally: } & \longrightarrow & \text{ Formally: }\\ \text{“A closed path in Y...} & \longrightarrow & \text{“A map } f: {\bf S}^1 \rightarrow Y ...\\ \text{can be deformed to...} & \longrightarrow & \text{is homotopic to...}\\ \text{a point.”} & \longrightarrow & \text{a constant function.”} \end{array}$$

Exercise. Indicate which of the following spaces are simply connected:

• disk: $\{(x,y) \in {\bf R}^2: x^2 + y^2 < 1 \}$;
• circle with pinched center: $\{(x,y) \in {\bf R}^2: 0 < x^2 + y^2 <1 \}$;
• ball with pinched center: $\{(x,y,z) \in {\bf R}^3: 0 < x^2 + y^2 + z^2 < 1 \}$;
• ring: $\{(x,y) \in {\bf R}^2: 1/2 <x^2 + y^2 <1 \}$;
• thick sphere: $\{(x,y,z) \in {\bf R}^3: 1/2 < x^2 + y^2 + z^2 < 1 \}$;
• the doughnut (solid torus): ${\bf T}^2 \times {\bf I}$.

Recall that the plane ${\bf R}^2$ is simply connected because every loop can be deformed to a point via a straight line homotopy:

More general is the following:

Theorem. Any convex subset of ${\bf R}^n$ is simply connected.

And so are all spaces homeomorphic to convex sets. There others too.

Theorem. The $n$-sphere ${\bf S}^n, \ n\ge 2$, is simply connected.

Proof. Idea for $n=2$. We assume that the loop $p$ isn't space-filling, so that there is a point $Q$ in its complement, $Q \in {\bf S}^2 \setminus \operatorname{Im}p.$ But ${\bf S}^2\setminus \{Q\}$ is homeomorphic to the plane! Since it also contains the loop, $p$ can be contracted to a point by a homotopy. $\blacksquare$

Exercise. Provide an explicit formula for the homotopy.

It is often more challenging to prove that a space is not simply connected. We will accept the following without proof (see Kinsey, Topology of Surfaces, p. 203).

Theorem. The circle ${\bf S}^1$ is not simply connected.

Corollary. The plane with a point taken out, ${\bf R}^2 \setminus \{(0,0) \}$, is not simply connected.

Exercise. Prove that corollary.

Simple-connectedness helps us classify manifolds.

Theorem (Poincare Conjecture). If $M$ is a simply connected compact path-connected $3$-manifold without boundary, then $M$ is homeomorphic to the $3$-sphere.

The Poincare sphere serves as an example that shows that simple-connectedness can't be replaced with $H_1=0$ for the theorem to hold.

Exercise. Prove that the sphere with a point taken out, ${\bf S}^2\setminus \{ N \}$, is simply connected.

Exercise. Prove that the sphere with two points taken out, ${\bf S}^2\setminus \{ N, S \}$, is not simply connected.

Exercise. Is the $3$-space with a line, such as the $z$-axis, taken out simply connected? Hint: Imagine looking down at the $xy$-plane:

Exercise. Is the $3$-space with a point taken out, ${\bf R}^3 \setminus \{(0,0,0)\}$, simply connected?

This analysis demonstrates that we can study the topology of a space $X$ indirectly, by studying the set of homotopy classes of maps $[Q,X]$ from a collection of wisely chosen topological spaces $Q$ to $X$. Typically, we use the $n$-spheres $Q={\bf S}^n,\ n=0,1,2,...$. These sets are called the homotopy groups and denoted by $$\pi_n (X):=[ {\bf S}^n, X].$$ Then, similarly to the homology groups $H_n(X)$, the sets $\pi_n (X)$ capture the topological features of $X$:

• $\pi_0(X)$ for cuts,
• $\pi_1(X)$ for tunnels,
• $\pi_2(X)$ for voids, etc.

In particular, we showed above that $\pi _1({\bf S}^n)$ is trivial for $n\ge 2$. These issues will be considered later in more detail.

## 4 Homotopy theory

We will seek a structure for the set of homotopy classes of maps.

First, what happens to two homotopic maps when they are composed with other maps? Are the compositions also homotopic?

In particular, if $f\simeq d:X\to Y$ and $q:X\to S$ is a map, are $qf,qg:x\to S$ homotopic too? As the picture below suggests, the answer is Yes:

The picture illustrates the right upper part of the following diagram:

$$\newcommand{\ral}[1]{\!\!\!\!\!\xrightarrow{\quad\quad\quad#1\quad\quad\quad}\!\!\!\!\!} \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} X & \ra{f\simeq g} & Y \\ \ \ua{p} & \searrow & \ \da{q} \\ R & \ra{pfq\simeq pgq} & S \end{array}$$

The result below refers to this diagram.

Theorem. Homotopy, as an equivalence of maps, is preserved under compositions. First, $$F:f\simeq g \Rightarrow H:qf\simeq qg,$$ where $$H(t,x) := qF(t,x),\ x \in X, t\in [0,1];$$ second, $$F:f\simeq g \Rightarrow H:fp\simeq gp,$$ where $$H(t,z) := F(t,p(z)),\ z \in R, t\in [0,1].$$

In either case, the new homotopy is the composition of the old homotopy and the new map.

Exercise. Finish the proof.

So, a map takes -- via compositions -- every pair of homotopic maps to a pair of maps that is also homotopic. Therefore, the map takes every homotopy class of maps to another homotopy class. Therefore, for any topological spaces $X,Y,Q$ and any map $$h:X\to Y,$$ the quotient map $$[h]:[Q,X] \to [Q,Y]$$ given by $$[h]([f]):=[hf]$$ is well defined.

In particular, $h:X\to Y$ generates a function on the homotopy groups: $$[h_n]:\pi_n(X) \to \pi_n(Y),\ n=0,1,2,...$$ The outcome is very similar to the way $h$ generates the homology maps: $$[h_n]:H_n(X) \to H_n(Y),\ n=0,1,2,...$$ as quotients of the maps of chains. However, in comparison, where is the algebra in these homotopy groups?

We need to define an algebraic operation on the homotopy classes. For example, given two homotopy classes of maps $f,g:{\bf S}^1 \to X$, i.e., two loops, what is the meaning of their “sum”, or their “product”, or some other combination? Under the homology theory approach, we'd deal with a formal sum of the loops. But such a "double" loop can't be a real loop, i.e., a map $f\cdot g:{\bf S}^1 \to X$.

Unless, the two loops have a point in common! Indeed, if $f(c)= g(c)$ for some $c\in {\bf S}^1$, then one can “concatenate” $f,g$ to create a new loop.

Exercise. Provide a formula for this construction.

Note that we are using the multiplicative notation for this group to be because, in general, it's not abelian.

Exercise. Give an example of two loops with $a\cdot b \not\simeq b \cdot a$.

It's similar for $n=2$. One can see these two intersecting (parametrized) spheres as one:

To make sure that this algebra works every time, we make a switch to the following.

Definition. Every topological space $X$ comes with a selected base point $x_0\in X$, denoted by $(X,x_0)$. They are called pointed spaces. Further, every map $f:(X,x_0)\to (Y,y_0)$ between two pointed spaces takes the base point to the base point: $f(x_0)=y_0$. These are called pointed maps.

Exercise. Prove that pointed spaces and pointed maps form a category.

This is, of course, just a special case of maps of pairs.

Now, the concatenation of two pointed maps $f,g:({\bf S}^n , u)\to (X,x_0)$ always makes sense as another pointed map $f\cdot g:({\bf S}^n , u)\to (X,x_0)$.

What about their homotopy classes? They are supposed to be the element of the homotopy groups. As usual, the product of the quotients is defined as the quotient of the product: $[f] \cdot [g] := [f\cdot g]$ for $f,g\in \pi_n(X,x_0)$. We will accept without proof that this operation is well defined and satisfies the axioms of group. The classes are with respect to the pointed homotopies, i.e., ones that remain fixed for the base points, as follows. Given two pointed maps $f_0, f_1: (X,x_0)\to (Y,y_0)$, a pointed homotopy is a homotopy $F: [0,1]\times X \to Y$ of $f_0, f_1$ which, in addition, is constant at $x_0$: for all $t\in [0,1]$ we have $$F(t,x_0) = f_0(x_0) = f_1(x_0).$$

Exercise. Prove that any pointed loop on a surface is homotopic to a loop that isn't onto. Derive that $\pi_1({\bf S}^2)=0$.

That's a group structure for the set of homotopy classes. Now a topology structure.

Recall that we have two interpretations of a homotopy $F$ between $f,g:X\to Y$. First, it's a continuous function ${\bf I}\times X \to Y$: $$F\in C({\bf I}\times X,Y).$$ Second, it's a continuously parametrized (by ${\bf I}$) collection of continuous functions $X\to Y$: $$F\in C({\bf I},C(X,Y)).$$ We conclude that $$C({\bf I}\times X,Y)=C({\bf I},C(X,Y)).$$ This is a particular case of the continuous version of the exponential identity of functions, as follows.

Proposition. For polyhedra $A,B,C$ we have $$C(A\times B,C)=C(A,C(B,C)).$$

Exercise. Prove the proposition.

We can take this one step further if the set of continuous functions $C(X,Y)$ is equipped with a topology. We can argue that this isn't just two sets with a bijection between them but a homeomorphism: $$C(A\times B,C)\approx C(A,C(B,C)).$$

Exercise. Suggest an appropriate choice of topology for this set. Hint: start with $A=B=C={\bf I}$.

From this point of view, homotopy is a path between two maps in the function space.

Exercise. What path-connected function spaces do you know?

## 5 Homotopy equivalence

Let's review what we've came up with in our quest for a relation between topological spaces that is “looser” than topological equivalence but still capable of detecting the topological features that we have been studying.

Spaces $X$ and $Y$ are called homotopy equivalent, or are of the same homotopy type, $X\simeq Y$, if there are maps $$f: X \to Y,$$ $$g: Y \to X$$ such that $$fg \simeq \operatorname{Id}_{Y},$$ $$gf \simeq \operatorname{Id}_{X}.$$ In that case, the notation is : $$f:X\simeq Y,\ g:Y\simeq X,$$ or simply: $$X\simeq Y.$$

Homotopy equivalence of two spaces is commonly described as a “deformation” and illustrated with a sequence of images that transforms one into the other:

Warning: in the common example above, the fact that the two spaces are also homeomorphic is irrelevant and may lead to confusion. To clarify, these as transformations preformed over a period of time:

• topological equivalence: continuous over space, incremental and reversible over time;
• homotopy equivalence: continuous over space, continuous and reversible over time.

Some of the examples of homotopies above also suggest examples of homotopy equivalent spaces.

For example, since all maps are homotopic here, the $n$-ball is homotopy equivalent to the point: $${\bf B}^n\simeq \{ 0\}.$$ Indeed, we can just contract the ball to its center. Note here that, if done incrementally, this contraction becomes a collapse and can't be reversed.

Exercise. Prove that any space homeomorphic to a convex subset of a Euclidean space is homotopy equivalent to a point.

Definition. A topological space $X$ is called contractible if $X$ is homotopy equivalent to a point, i.e., $X\simeq \{x_0\}$.

Example. A less obvious example of a contractible space is “the two-room house”. Here we have two rooms each taking the whole floor. There is access to the first floor through a tube from the top and to the second floor from the bottom:

These are the steps of the deformation:

• 1. we expand the entries into the building at the top and bottom to the size of the whole circle, which turns the two tubes into funnels;
• 2. then we expand the entries into the rooms from the funnels to the half of the whole circle creating a sloped ellipse;
• 3. finally, we contract the walls and we are left with only the ellipse.

$\square$

Exercise. Prove that the dunce hat is contractible:

Further, our intuition suggests that “thickening” of a topological space doesn't introduce new topological features. Indeed, we have the following.

Theorem. For any topological space $Y$, $$Y\times {\bf I} \simeq Y.$$

Proof. The plan of the proof comes from our analysis of circle vs. cylinder. Let $$X={\bf I} \times Y.$$ Consider the two natural maps. The first is the projection $$p:X\to Y,\ p(t,x)=x,$$ of $X$ onto $Y$ as its “bottom” and the second is the embedding $$e:Y\to X, \ e(x)=(0,x)$$ of $Y$ into $X$ as its “bottom”.

Let's consider their compositions:

• $pe:Y\to Y=\operatorname{Id}_Y$;
• $ep:X\to X$ is the collapse:

$$ep(t,x)=(0,x).$$ For the latter, we define the homotopy $$H:[0,1]\times [0,1]\times Y \to [0,1]\times Y$$ by $$H(t,s,x):=(ts,x).$$ It is easy to confirm that: $$H(0,\cdot)=ep, \ H(1,\cdot)=\operatorname{Id}_X.$$

Next, we prove that this is a continuous map with respect to the product topology. Suppose $U$ is an element of the standard basis of the product topology of $[0,1]\times Y$, i.e., $U=(a,b)\times V$, where $a < b$ and $V$ is open in $Y$. Then \begin{align} H^{-1}(U)&=H^{-1}((a,b)\times V) \\ &= \{ (t,s)\in [0,1]\times [0,1]: a < ts < b \} \times V \end{align} The first component of this set is a region between two hyperbolas $s=a/t,s=b/t$. It is open and so is $H^{-1}(U)$.

$\blacksquare$

The main properties are below.

The first two are obvious:

Theorem.

• Homotopy invariance is a topological property.
• A homeomorphism is a homotopy equivalence.

Later we will show the following:

• Homology groups are preserved under homotopy equivalence.

Theorem. Homotopy equivalence is an equivalence relation for topological spaces.

Proof. 1. Reflexivity: $f,g:X\simeq X$ with $f=g=\operatorname{Id}_X$.

2. Symmetry: if $f,g:X\simeq Y$ then $g,f:Y\simeq X$.

3. Transitivity: if $f,g:X\simeq Y$ and $p,q:Y\simeq Z$ then $pf,gq:X\simeq Z$. Indeed: $$(pf)(gq)=p(fg)q \simeq p \operatorname{Id}_X q = pq \simeq \operatorname{Id}_Z,$$ $$(gq)(pf)=g(qp)f \simeq g \operatorname{Id}_Z f = gf \simeq \operatorname{Id}_X,$$ by the theorem from the last subsection.

$\blacksquare$

Once again, homotopy helps us classify manifolds.

Theorem (Generalized Poincaré Conjecture). Every compact $n$-manifold without boundary which is homotopy equivalent to the $n$-sphere is homeomorphic to the $n$-sphere.

Exercise. (a) Prove that the above spaces are homotopy equivalent but not homeomorphic. (b) Consider other versions of the square with a handle and find out whether they are homeomorphic and/or homotopy equivalent. (c) What about the Mobius band with a handle? (d) What about the sphere with a handle? Hint: don't keep the sphere in $3$-space.

Exercise. Prove that ${\bf S}^1 \times {\bf S}^1$ and ${\bf S}^3 \vee {\bf S}^1$ are not homotopy equivalent.

Exercise. Show that ${\bf S}^2 \vee {\bf S}^1 \simeq {\bf S}^2 \cup A$, where $A$ is the line segment from the north to south pole.

Exercise. These spaces are homotopy equivalent to some familiar ones: (a) ${\bf S}^2 \cup A$, where $A$ is disk bounded by the equator, (b) ${\bf T}^2 \cup D_1 \cup D_2$, where $D_1$ is a disk bounded by the inner equator and $D_2$ is a disk bounded by a meridian.

Exercise. With amoeba-like abilities, can this person unlink his fingers without unlocking them? Hint: worry about the shirt later.

Reminder: Homotopy is a relation between maps while homotopy equivalence is a relation between spaces.

## 6 Homotopy in calculus

The importance of these concepts can be seen in calculus, as follows.

Recall a vector field in the plane is a function $V: {\bf R}^2 \to {\bf R}^2$. It is called conservative on a region $D$ if it is the gradient of a scalar function: $$V = \nabla f.$$ Such a vector field may represent the vertical velocity of a flow on a surface ($z=f(x,y)$) under gravity or a force of a physical systems in which energy is conserved.

We know the following.

Theorem. Suppose we have a vector field $V = (P,Q)$ defined on an open set $D \subset {\bf R}^2$. Suppose $V$ is irrotational: $$P_y = Q_x.$$ Then $V$ is conservative provided $D$ is simply connected.

It is then easy to prove then that the line integral along a closed path is $0$.

But what if the region isn't simply connected? The theorem doesn't apply anymore but, with the tools of this section, we can salvage a lot.

Theorem. Suppose we have an irrotational vector field $V = (P,Q)$ defined on an open set $D \subset {\bf R}^2$. Then the line integral along any closed path homotopic to a constant, called null-homotopic, is $0$.

The further conclusion is that line integrals are path-independent, i.e., the choice of integration path from a point to another does not affect the integral.

The issue of path-independence is related to the following question: What if we can get from point $a$ to point $b$ in two “topologically different” ways?

In other words, there are several homotopy classes of paths from $a$ to $b$ relative to the end-points.

Exercise. Prove from the theorem that the line integrals are constant within each of these homotopy classes.

Recall next for any continuous vector field in ${\bf R}^n$, the forward propagation map $$Q_t:{\bf R}^n\to {\bf R}^n$$ is given by $$Q_t(a)=a+tV(a).$$ Of course, any restriction of $Q_t\Big|_{R}, \ R\subset {\bf R}^n$, is also continuous. Suppose $R$ is given and we'll use $Q_t$ for this restriction.

Now we observe that it is continuous with respect to either of the variables.

We realize that this is a homotopy!

But not just any; we have: $$Q_0(a)=a, \ \forall a \in R.$$ In other words, this is the inclusion $i_R:R\hookrightarrow {\bf R}^n$, $$Q_0=i_R.$$

Then we have the following.

Theorem. The forward propagation map $Q_t:R \to {\bf R}^n, \ t>0,$ generated by a continuous vector field is homotopic to the inclusion.

Corollary. Suppose the forward propagation map $Q_t:R \to S$ is well defined for some $S\subset {\bf R}^n$, and some $t>0$. Suppose also that $R$ is a deformation retract of $S$. Then we have: $$rQ_t \simeq \operatorname{Id}_R,$$ where $r:S\to R$ is the retraction.

Of course, when $S=R$, we have simply $$Q_t \simeq \operatorname{Id}_R.$$

Therefore, as we shall see later, the possible behaviors of dynamical systems defined on a space is restricted by its topology.

Exercise. Prove the corollary.

Theorem (Fundamental Theorem of Algebra). Every non-constant (complex) polynomial has a root.

Proof. Choose the polynomial $p$ of degree $n$ to have the leading coefficient $1$. Suppose $p(z)\ne 0$ for all $z\in {\bf C}$. Define a map, for each $t>0$, $$p_t:{\bf S}^1 \to {\bf S}^1,\quad p_t(z):=\frac{p(tz)}{|p(tz)|}.$$ Then $p_t\simeq p_0$ for all $t>0$ with $p_0$ a constant map. This contradicts the fact that $p_t \simeq z^n$ for large enough $t$. $\blacksquare$

Exercise. Provide the details of the proof.

## 7 Is there a constant homotopy between constant maps?

We know that path-connectedness implies that every two constant maps are homotopic and homotopic via a “constant” homotopy, i.e., one with every intermediate function $f_t$ constant. What about the converse?

Suppose $f_0,f_1:X\to Y$ are two constant maps. If they are homotopic, does this mean that there is a constant homotopy between them?

You can fit the short version of the solution, if you know it, in this little box: $$[\quad\quad\quad\quad\quad]$$

However, what if we don't see the solution yet? Let's try to imagine how we would discover the solution following the idea: "Teach a man to fish..."

Let's first re-write the problem in a more concrete way:

• given $f_0,f_1:X\to Y$, that satisfy
• $f_0(x)=c\in Y,\ f_1(x)=d\in Y$ for all $x\in X$;
• also given a homotopy $F:f_0\simeq f_1$, where $F:[0,1]\times X\to Y$, with $F(t,\cdot)=f_t$.

Is there a homotopy $G:f_0\simeq f_1$, where $G:[0,1]\times X\to Y$, with $G(0,\cdot)=f_0,G(1,\cdot)=f_1$ and $G(t,x)=g(t)$ for all $t,x$?

We will attempt to simplify or even “trivialize” the set-up until the solution becomes obvious, but not too obvious.

The simplest setup appears to be a map from an interval to an interval. Below we show a constant, $G$, and a non-constant, $F$, homotopies between two constant maps:

Of course, this is too narrow as any two maps are homotopic under these circumstances: $Y$ is convex. The target space is too simple!

To get around this obviousness, we should

• try to imagine that $Y$ is slightly bent, and then
• try to build $G$ directly from $F$.

But, really, how do we straighten out these waves?

The fact that this question seems too challenging indicates that the domain space is too complex!

But what is the simplest domain? A single point!

In this case, all homotopies are constant. The domain space is too simple!

What is the next simplest domain? Two points!

Let's make the setup more concrete:

1. Given $f_0,f_1:\{a,b\}\to [c,d]$ (bent),
2. $f_0(a)=f_0(b)=c,\ f_1(a)=f_1(b)=d$;
3. also given a homotopy $F:f_0\simeq f_1$, where $F:[0,1]\times \{a,b\}\to [c,d]$, with $F(t,\cdot)=f_t$.
4. Find a homotopy $G:f_0\simeq f_1$, where $G:[0,1]\times \{a,b\}\to [c,d]$, with $G(0,\cdot)=f_0,\ G(1,\cdot)=f_1$ and $G(t,a)=G(t,b)$ for all $t$.

The setup is as simple as possible. As the picture on the left shows, everything is happening within these two green segments.

Meanwhile, the picture on the right is a possible homotopy $F$ and we can see that it might be non-trivial: within either segment, $d$ is “moved” to $c$ but at a different pace.

The setup is now very simple but the solution isn't obvious yet. That's what we want for our analysis!

Items 1 and 2 tell us that there are just two values, $c$ and $d$, here. Item 3 indicates that there are just two functions, $F(\cdot,a)$ and $F(\cdot,b)$ here, and item 4 asks for just one function, $G(\cdot,a)$ same as $G(\cdot,b)$.

How do we get one function from two?!

It is tempting to try to combine them algebraically (e.g., via a convex combination) but remember, $[c,d]$ is bent and there is no algebra.

So, we need to construct $G(\cdot,a)=G(\cdot,b)$ from $F(\cdot,a)$ and $F(\cdot,b)$.

What to do?..

Exercise. Fill the box: $$[\quad\quad\quad\quad\quad]$$

## 8 Homotopy equivalence via cell collapses

The idea of homotopy equivalence allows one try to simplify a topological space before approaching its analysis. This idea seems good on paper but, practically, how does one simplify a given topological space?

A stronger version of the Nerve Theorem, which we accept without proof (see Alexandroff and Hopf Topology) gives us a starting point.

Theorem (Nerve Theorem). Let $K$ be a (finite) simplicial complex and $S$ an open cover of its realization $|K|$. Suppose the finite intersections of the elements of $S$ are contractible. Then the realization of the nerve of $S$ is homotopy equivalent to $|K|$.

The question becomes, how does one simplify a given simplicial complex $K$ while staying within its homotopy equivalence class?

We shrink cells one at a time:

As we see, we have reduced the original complex to one with just seven edges. Its homology is obvious.

Now, more specifically, we remove cell $\sigma$ and one of its boundary cells $a$ at every step, by gradually pulling $\sigma$ toward the closure of $\partial \sigma \setminus a$. Two examples, $\dim \sigma =1$ and $\dim \sigma =2$, are given below:

This step, defined below, is called an elementary collapse of $K$.

Of course, some cells can't be collapsed depending on their place in the complex:

The collapsible $n$-cells are marked in orange, $n=1,2$. As we see, for an $n$-cell $\sigma$ to be collapsible, one of its boundary $(n-1)$-cells, $a$, has to be a “free” cell in the sense that it is not a part of the boundary of any other $n$-cell. The free cells are marked in purple. The rest of the $n$-cells aren't collapsible because each of them is either fully surrounded by other $n$-cells or is in the boundary of an $(n+1)$-cell.

Let's make this construction more precise.

The shrinking is understood as a homotopy equivalence -- of the cell and, if done correctly, of the whole complex. We deform the chosen cell to a part of its boundary and do that in such a way that the rest of the complex remains intact!

The goal is accomplished via a more general type of homotopy than the pointed homotopy.

Definition. Given two maps $f_0, f_1: X \to Y$ and a subset $A \subset X$, the maps are homotopic relative to $A$ if there is a homotopy $F: [0,1]\times X \to Y$ of $f_0, f_1$ which, in addition, is constant on $A$: for all $t\in [0,1]$ we have $$F(t,a) = f_0(a) = f_1(a)$$ for all $a \in A.$ We use the following notation: $$f_0 \simeq f_1 \quad {\rm rel } \ A.$$

Theorem. Suppose $\sigma$ is a geometric $n$-simplex and $a$ is one of its $(n-1)$-faces. Then $\sigma$ is homotopy equivalent to the union of the rest of its faces relative to this union: $$\sigma \simeq \partial \sigma \setminus a \quad {\rm rel } \ \partial \sigma \setminus a.$$

Exercise. Provide a formula for this homotopy equivalence. Hint: use barycentric coordinates.

Definition. Given a simplicial complex $K$, suppose $\sigma$ is one of its $n$-simplices. Then $\sigma$ is called a maximal simplex if it is not a boundary simplex of any $(n+1)$-simplex. A boundary $(n-1)$-simplex $a$ of $\sigma$ is called a free face of $\sigma$ if $a$ is not a boundary simplex of any other $n$-simplex.

In other words, $a$ is a maximal simplex in $K\setminus \sigma$.

Proposition. If $\sigma$ is a maximal simplex in complex $K$ and $a$ is its free face, then $K_1=K\setminus \{\sigma,a\}$ is also a simplicial complex.

Exercise. Prove the proposition.

This removal of two cells from a simplicial complex is called an elementary collapse and the step is recorded with the following notation: $$K \searrow K_1:=K\setminus \{\sigma, a\}.$$

Corollary. The homotopy equivalence of a realization of a simplicial complex is preserved under elementary collapses.

Definition. A simplicial complex $K$ is called collapsible if there is a sequence of elementary collapses of $K$ that ends at a single vertex $V$: $$K \searrow K_1 \searrow K_2 \ ... \ \searrow K_N =\{V\}.$$

Corollary. The realization of a collapsible complex is contractible.

Exercise. Show that the complexes of the dunce hat and the two-room house are examples of contractible but not collapsible complexes.

Since it is presented in the language of cells, the above analysis equally applies to both simplicial and cubical complexes.

Exercise. Prove the theorem for cubical complexes.

Exercise. Define the “non-elementary” cell collapse that occurs when, for example, a triangle has two free edges and the total of four cells are removed. Repeat the above analysis. Hint: It is the vertex that is free.

Exercise. Define elementary expansions as the “inverses” of elementary collapses and repeat the above analysis.

Exercise. Prove that the torus with a point taken out is homotopy equivalent to the figure eight: ${\bf T}^2 \setminus \{u\} \simeq {\bf S}^1 \vee {\bf S}^1$.

## 9 Invariance of homology under cell collapses

The real benefit of this construction comes from our realization that the homology can't change. In fact, later we will show that the homology groups are always preserved under homotopy equivalence. With cell collapses, the conclusion is intuitively plausible since it seems impossible that new topological features could appear or the existing features could disappear.

Let's analyze what we've got. Suppose we have a simplicial complex $K$ and a collapse: $$K \searrow K^1 .$$ We suppose that from $K$ an $n$-cell $\sigma$ and its free face $a$ are removed.

We need to find an isomorphism on the homology groups of these two complexes. We don't want to build it from scratch but instead find a map that induces it. Of course, we choose the inclusion $$i:K^1 \hookrightarrow K.$$ Since this is a simplicial map, it induces a chain map: $$i_{\Delta}:C(K^1)\to C(K),$$ which is also an inclusion. This chain map generates a homology map: $$i_*:H(K^1)\to H(K).$$ We expect it to be an isomorphism.

Exercise. Define the projection $P:C(K)\to C(K^1)$. What homology map does it generate?

For brevity, we denote: $$\partial:=\partial^{K},\ \partial^1:=\partial^{K^1}.$$ This is what the chain complexes and the chain maps look like: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccc} ...& \ra{\partial_{n+1}} & C_{n}(K) & \ra{\partial_{n}} & C_{n-1}(K)& \ra{\partial_{n-1}} &...\\ ...& & \ua{i_n} & & \ua{i_{n-1}}& &...\\ ...& \ra{\partial_{n+1}^{1}} & C_{n}(K^1) & \ra{\partial_{n}^{1}} & C_{n-1}(K^1)& \ra{\partial_{n-1}^{1}} &... \end{array}$$ Only the part of this chain map diagram is affected by the presence or absence of these two extra cells. The rest has identical rows and columns: $$C_k(K)=C_k(K^1),\ \forall k \ne n,n-1,$$ $$\partial_{k} = \partial_{k}^1,\ \forall k\ne n+1,n,n-1,$$ $$i_k=Id_{C_k(K^1)},\ \forall k \ne n,n-1.$$

Let's compute the specific values for all elements of the diagram.

These are the chain groups and the chain maps: $$\begin{array}{lll} C_n(K) &=C_n(K^1)\oplus < \sigma >\\ C_{n-1}(K) &=C_{n-1}(K^1)\oplus < a >,\\ i_n &=Id_{C_n(K^1)}\oplus 0,\\ i_{n-1} &=Id_{C_{n-1}(K^1)}\oplus 0. \end{array}$$

Now, we need to express $\partial_k$ in terms of $\partial_k^1$. Then, with the boundary operators given by their matrices, we can handle cycles and boundaries in a purely algebraic way...

For the boundary operator for dimension $n+1$, the matrix's last row is $\partial^{-1} \sigma$ and it is all zero because $\sigma$ is a maximal cell: $$\partial_{n+1}=\partial_{n+1}^1 \oplus 0= \left[ \begin{array}{ccccccccccccccc} \\ \\ & \partial_{n+1}^1 \\ \hline \\ \\ 0,0,&... &,0 \end{array} \right]:C_{n+1}(K^1) \to C_{n}(K^1)\oplus < \sigma >.$$ Then $$\ker \partial _{n+1}=\ker \partial _{n+1}^1 ,\operatorname{Im} \partial _{n+1}=\operatorname{Im} \partial _{n+1}^1 \oplus 0.$$

For the boundary operator for dimension $n$, the matrix's last row is $\partial^{-1} a$ and it has only one $1$ because $a$ is a part of the boundary of only one cell. That cell is $\sigma$ and the last column, which is non-zero, is $\partial \sigma$ with $1$s corresponding to its faces: $$\partial_{n}=\left[ \begin{array}{cccc|c} &&&&\pm 1\\ &&&&\pm 1\\ &&\partial_{n}^1&&0\\ &&&&\vdots\\ &&&&0\\ \hline 0,&0,&... &,0&\pm 1 \end{array} \right]:C_{n}(K^1)\oplus < \sigma > \to C_{n-1}(K^1)\oplus < a >.$$ Then $$\ker \partial _{n}=\ker \partial _{n}^1 \oplus 0 ,\operatorname{Im} \partial _{n}=\operatorname{Im} \partial _{n}^1 \oplus \partial \sigma.$$

For the boundary operator for dimension $n-1$, the matrix's last column is $\partial a$, which is non-zero, with $1$s corresponding to $a$'s faces: $$\partial_{n-1}=\left[ \begin{array}{cccc|c} &&&&\pm 1\\ &&&&\pm 1\\ &&\partial_{n-1}^1&&0\\ &&&&\vdots\\ &&&&0 \end{array} \right]:C_{n-1}(K^1)\oplus < a > \to C_{n-2}(K^1).$$ Then $$\ker \partial _{n-1}=\ker \partial _{n-1}^1 \oplus 0 ,\operatorname{Im} \partial _{n-1}=\operatorname{Im} \partial _{n-1}^1 .$$

Finally, the moment of truth... $$H_{n+1}(K) := \frac{\ker \partial _{n+1}}{\operatorname{Im} \partial _{n+2}} = \frac{\ker \partial _{n+1}^1 }{\operatorname{Im} \partial _{n+2}^1} =: H_{n+1}(K^1);$$ $$H_n(K) := \frac{\ker \partial _{n}}{\operatorname{Im} \partial _{n+1}} = \frac{\ker \partial _{n}^1 \oplus 0}{\operatorname{Im} \partial _{n+1}^1 \oplus 0} \cong \frac{\ker \partial _{n}^1}{\operatorname{Im} \partial _{n+1}^1 }=: H_n(K^1);$$ $$H_{n-1}(K) := \frac{\ker \partial _{n-1}}{\operatorname{Im} \partial _{n}} = \frac{\ker \partial _{n-1}^1 \oplus 0}{\operatorname{Im} \partial _{n}^1} \cong \frac{\ker \partial _{n-1}^1}{\operatorname{Im} \partial _{n}^1 }=: H_{n-1}(K^1).$$ The rest of the homology groups are unaffected by the two extra cells: $$H_{k}(K)\cong H_{k}(K^1),\forall k\ne n,n-1,n-2.$$

We have proven the following.

Theorem. If $K \searrow K^1$ then $$H(K)\cong H(K^1),$$ under the homology map induced by the inclusion $i:K_1\hookrightarrow K$.

Exercise. Provide details for the last part of the proof.

All these lengthy computations were needed to demonstrate something that may seem obvious, that these spaces have the same homology:

Exercise. Suppose $K$ is a $2$-dimensional simplicial complex. Suppose $\sigma$ is a $2$-cell of $K$ that has two free edges, $a,b$, with vertex $A$ between them. Let $K^1:=K\setminus \{\sigma, a,b,A\}$. Use the above approach to prove that $H(K)\cong H(K^1)$.

Exercise. Under what conditions is an elementary collapse a homeomorphism?