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From Intelligent Perception

The issue to be discussed is:

By a "map" we understand a continuous function but the word "transformed" will have to be explained. We are interested in the possibility of gradually and continuously transforming two maps, between the same spaces, into each other. First, "gradually" indicates the need for intermediate maps between the two and, second, those maps are expected to be continuous.

Sometimes things are simple. Any map $f:X \rightarrow {\bf R}$ can be "transformed" into any other map $g$.

Indeed, one can "gradually" slide $f(x)$ toward $g(x)$. In fact it's easy to think of a simple formula: $$F(t,x) = (1-t)f(x) + tg(x),$$ where $t$ is "time".

The same exact formula applies if ${\bf R}$ is replaced with ${\bf R}^n$ (or any convex set in a topological vector space).

Such a transformation is called homotopy.

Let' consider maps of the circle to the ring:

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

The first four bands (and maps) can be transformed into each other but not the last one. Why? How do we even know?

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

To get from $f_0$ (the smaller circle) to $f_0$ (the large circle), one goes through intermediate step - 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.

Definition. Two maps $f_0, f_1: X → Y$ are called homotopic if there is a map $F: [0,1]×X → Y$ such that $$F(0,x) = f_0(x), F(1,x) = f_1(x)$$ for all $x∈X.$ Map $F$ is called a homotopy.

Notation: $$f_0 \sim f_1.$$ Indeed:

Theorem. Homotopy is an equivalence relation.

Proof. Proving reflexivity and symmetry is easy. For transitivity one needs to figure out this: $$f \sim ^F g \sim ^G h \Rightarrow f \sim ^? h.$$ Exercise. $\blacksquare$

Exercise. What can you say about the homotopies of this map?

Also important is the relative homotopy. Given two maps $f_0, f_1: X → Y$ and a subset $A \subset X$, then the maps are homotopic relative to $A$ if there is a homotopy $F: [0,1]×X → Y$ of $f_0, f_1$ which, in addition, is constant on $A$: $$F(t,a) = f_0(a) = f_1(a)$$ for all $a∈A.$ Notation: $f_0 \sim f_1$ rel $A$.

Sidenote: One defines the fundamental group via relative homotopy of loops.

Sidenote: Homology classes are preserved under homotopy, see homology of homotopic maps.

Notation: The homotopy may be specified: $$f_0 \sim ^F f_1.$$

Theorem. Homotopy is preserved under compositions:

• $f\sim ^Fg \Rightarrow hf\sim ^Hhg$ with $H(t,x)=hF(t,x), x \in X$;
• $f\sim ^Fg \Rightarrow fh\sim ^Hgh$ with $H(t,z) = F(t,h(z)),z \in Z$.

The set-up of the theorem is illustrated by the following diagram: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccc} X & & Z \\ \da{f} \da{g} & & \da{h} \\ Y & & X \\ _{ }\da{h} & & \da{f} \da{g} \\ Z & & Y \end{array} $$

We call $R \subset {\bf R}^n$ simply connected if every closed path can be deformed to a point.

But what does "deformed" mean? We will use the notion of homotopy to make this precise.

We translate:

Briefly, region $R$ is simply connected if every closed path is homotopic to a constant path.

Theorem. The plane ${\bf R}^2$ is simply connected.

Proof. There is a required homotopy:

Let's find a formula.

Let's define $$P \colon [0,1] \times {\bf R}^2 \rightarrow {\bf R}^2$$ with this formula: $$P(t,u) = (1-t)u.$$ This transformation contracts the whole plane to the origin. Now we just choose: $$F(t,s)=P(t,f_1(s)).$$ That's our homotopy. $\blacksquare$

The same proof applies to:

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

Exercise: Prove that the plane with a point taken out, ${\bf R}^2 - \{(0,0)\}$, is not simply connected. Hint: start with $(\sin s, \cos s)$ and continue by contradiction.

Example: As we know any two continuous functions $f, g \colon [a,b] \rightarrow {\bf R}^n$ are homotopic. We can use $P$ as a homotopy between $P(0,u) = u$, the identity map and $P(1,u) = 0$, the constant map.

Define $F \colon [0,1] \times [0,1] \rightarrow {\bf R}^2$ so that $F(0,s) = f_0(s)$ and $F(1,s) = f_1(s)$ by $F(t,s) = P(t, f_0(s))$ (continuous as a composition). Verify:

• $F(0,s) = P(0,f_0(s)) = f_0(s)$ and
• $F(1,s) = P(1,f_0(s)) = 0 = f_1(s)$.

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

Proof. See under Simple connectedness. $\blacksquare$

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

Exercise. Prove that the sphere ${\bf S}^2$ is simply connected:

Hint: use the fact that the ball ${\bf B}^3$ is simply connected.

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

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

Exercise. Prove that the $3$-space with a point taken out, ${\bf R}^3 \backslash \{(0,0,0)\}$, is simply connected.

Exercise. Is the $3$-space with a line taken out, ${\bf R}^3 \backslash \{{\rm z-axis} \}$, simply connected?

Hint: Imagine looking down at the $xy$-plane.

Theorem. The torus ${\bf T}^2$ is not simply connected:

To justify the interpretation of homotopy as a continuous deformation we prove the following:

Theorem. For any homotopy $F:[0,1]\times X\rightarrow Y$ and any $t\in [0,1]$ the map $$f_t(\cdot)=F(t,\cdot):X \rightarrow Y$$ is continuous.

Proof. It follows from the fact that a restriction of a continuous function is continuous. $\blacksquare$

This analysis suggest another interpretation of homotopy -- via the exponential identity of functions: $$[A\times B,C]=[A,[B,C]],$$ or, in our case, $$[{\bf I}\times X,Y]=[{\bf I},[X,Y]],$$ where ${\bf I}=[0,1]$. In other words, a homotopy is a function parametrized by the numbers in ${\bf I}$.

Then, indeed, not only a homotopy is a function: $$F:{\bf I}\times X\rightarrow Y$$ but also $$F:{\bf I}\rightarrow [X,Y].$$ In fact, since $F$ has to be continuous, a more complex identity is required: $$C(A\times B,C)=C(A,C(B,C)),$$ and the space of continuous functions $C(B,C)$ will have to be equipped with a topological structure assigned.

Exercise. Suppose $f_0,f_1:X\rightarrow Y$ are two constant maps. If they are homotopic, does this mean that there is a "constant" homotopy, i.e., the one with every function $f_t$ constant, between them? What does this have to do with path-connectedness? Solution.

See also Homotopy as a perturbation.