Taking Homotopy to the next level

In the last post I’ve presented the term of homotopy, defined the homotopy category and homotopic equivalent spaces. In this post I would like to present some more properties and definitions related to homotopy and get some results. Just one note before I start: I will only deal with continuous maps from now on – Unless I say otherwise

Contractible spaces

I’ll start with a definition: Let X,Y be two topological spaces, and suppose that f:X\to Y is a (continuous) map between them. We say that f is null-homotopic if it is homotpic to a constant map.

Intuitivly, it means that you can continuously transform f into a constant function. Let’s see an example: Pick a map f:X\to \mathbb{R}^n. Then the function:

H:X\times I\to \mathbb{R}^n,H(x,t)=(1-t)f(t)

is indeed a homotopy between f to the constant function K_0(x)=0. Thus, every function into \mathbb{R}^n is null-homotopic.

With the term of a null-homotopic map we can define what a contractible space is: We say that X is a contractible space if Id_X is null-homotopic. Using the example in the above, we can subsitute X=\mathbb{R}^n,f=Id_{\mathbb{R}^n} and conclude that \mathbb{R}^n is null homotopic.

I want to give some more intuition to what a contractible space is using the following statement:

X\text{ is contractible}\iff X\text{ is homotopic equivalent to a point}

This statment allows us to think of X as a space that we can continuously shrink to a single point.

Shrinking \mathbb{R}^2 to the origin

Let’s prove it: Suppose that X is contractible and \{p\} is a space with a single point, we know that, Id_X is homotopic to a constant function K_a (where K_a(x)=a for every x\in X).

Let’s define a map f:X\to \{p\} where f(x) = p (we actually don’t really have that much of a choice here…). We need to prove that f is an homotopic equivalence. i.e. we need to find a map g:\{p\}\to X such that:

f\circ g\sim Id_{\{p\}},g\circ f\sim Id_{X}

I’ll pick the map g(p)=a. Let’s check if it valid:

f\circ g(p)=f(g(p))=f(a)=p\Longrightarrow f\circ g = Id_{\{p\}}\Longrightarrow f\circ g\sim Id_{\{p\}}

Great, now on the other hand:

g\circ f(x) = g(f(x))=g(p)=a\Longrightarrow g\circ f\sim K_a

And by our assumption, K_a\sim Id_X, and since \sim is transitive, we get that g\circ f \sim Id_X. And we have proved that f is a homotopic equivalence, as we wanted.

Now, suppose that X is homotopic equivalent to a single point space \{p\}. Therefore, the function f:X\to\{p\} is an homotopic equivalence (since it is the only map from X to \{p\}). Therefore, there is a map g:\{p\}\to X such that:

f\circ g\sim Id_{\{p\}},g\circ f\sim Id_{X}

Let a:= g(p)\in X, so:

g\circ f(x)=g(f(x))=g(p)=a\Longrightarrow g\circ f = K_a

And we know that g\circ f \sim Id_X, thus K_a \circ Id_X so by definition, X is contractible.

Homotopy with respect to subspace

Great, we now have a nice intuition to contractible space, this property will be important to us later. I want to define now a new kind of homotopy: Given two topological spaces X,Y, a subspace A\subseteq X and two maps f,g:X\to Y, we will say that f is homotopic to g with respect to A, and denote f\sim_{A}g if there is a homotopy H:X\times I\to Y from f to g if H(a,0)=H(a,t) for every a\in A,t\in I. There isn’t really anything new here, this is just a regular homotopy with some restrictions, the value of elements of the set A shall stay the same at any given point at time (t\in I). There is a nice animation in the wikipedia of such an homotopy:

Those are two paths on the plane (\gamma : [0,1]\to \mathbb{R}^2) which are homotopic equivalent with respect to the set \partial [0,1] = \{0,1\}.

It’s not hard to see that \sim_A is an equivalnce relation and a necessary condition for f and g to be homotopy equivalent with respect to A is: f|_A=g|_A. This kind of homotopy will be important later, so remember it!


I am now going to talk about special case of a subset, I am going to use the inclusion map: i:A\to X where A\subseteq X and i(a)=a.

As always, X is a topological space and A\subseteq X. We will say that A is a retract of X if there is a map r:X\to A such that r\circ i = Id_A.

Let’s try to understand what we are facing here. We can think about retract as a subspace that we can continuously ‘squish’ the whole space into, while not moving the the subspace itself. It kind of makes sense to think of A as a dominant subspace that can ‘carry on his back’ the whole space.

I don’ t if you noticed, but I really like demonstrating thing in topology with animations, so I’ll give an example Consider the “8” space:

(Formally, It is a qutient space of two intervals where we identify the endpoints as the same point), The lower / upper circles are both retracts. I’ll show here two ways to map the space into the lower circle, by ‘shrinking’ the upper circle or by ‘folding’ the space.

A topolical space and it’s retract have some shared properties, for example, if X is contractible, and A\subseteq X is a retract, then A is contractible as well. Let’s prove it:

By definiton, we know that Id_X\sim K_p for some p\in X. Since A is a retract, there is a map r:X\to A such that r\circ i = Id_A. Suppose that H: X\times I \to X is a homotopy from Id_X to K_p. Let’s take now the map:

r\circ H\circ (i\times Id_{[0,1]}):A\times I\overset{i\times Id_{[0,1]}}{\longrightarrow}X\times I\overset{H}{\longrightarrow}X\overset{r}{\longrightarrow}A

It is continuous as a composition of continuous maps, and:

r\circ H\circ (i\times Id_{[0,1]})(a,0)=r\circ H((i\times Id_{[0,1]})(a,0))
=r\circ H(a,0)=r\circ Id_X(a)=r(a)=a

Thus, r\circ H\circ (i\times Id_{[0,1]}) = Id_A (notice that r(a) = a since r\circ i = Id_A). Moreover:

r\circ H\circ (i\times Id_{[0,1]})(a,1)=r\circ H((i\times Id_{[0,1]})(a,1))
=r\circ H(a,1)=r\circ K_p(a)=r(p)\in A

Thus, r\circ H\circ (i\times Id_{[0,1]})(a,1) = K_{r(p)} and we proved that Id_A\sim K_{r(p)}, therefore, A is contractable, as we wanted.

Now I Want to give a counter-exmaple: Pick X=[0,1] and A=\partial [0,1] = \{0,1\}. I state that A is not a retract. Suppose it is, then there is a continuous map r:[0,1]\to{0,1} such that r(0)=0,r(1)=1, r(x)\in\{0,1\}, we can exapnd this function’s image to the whole interval by defining: r^\prime:[0,1]\to [0,1], r^\prime(x)=r(x). And now we have a contradiction to the Intermediate value theorem (think why!).

I want to present now a stronger type of retract: Suppose that X is a topological space, A\subseteq X. We say that A is a deformation retract of X if there is a map r:X\to A such that:

  • r\circ i = Id_A (nothing new here)
  • i\circ r \sim_A Id_X

This is a stronger property, we are not only asking if such a map r exists, we also want r to be a homotopy equivalence with respect to A, therefore, A is homotopic equivalent to X with respect to A.

Notice that in order to check that a retract is a deformation retract we only need to find a homotopy with respect to A between i\circ r and the identity on X.

As you may guess, it is now time to give an example. I’ll pick \mathbb{R}^n-\{0\} and:


We need to define a homotopy:

H:(\mathbb{R}^n-\{0\})\times I\to (\mathbb{R}^n-\{0\})

such that H(x,0)=Id_X,H(x,1)=i\circ r. Wait a minute, what is r? I’ll use the fact that the points in S^{n-1} are exactly the points with norm one, so let’s take the function that normalizes it’s input: r(x)=\frac{x}{||x||}. This function indeed satisfies the first property, and we can now define the homotopy:

H(x,t)=(1-t)\cdot x+t\cdot\frac{x}{||x||}

That is indeed a homotopy with respect to A, thus S^{n-1} is a deformation retract.

I don’t know about you, but I really like this animation, and it seems like good place to stop.


I’ve shown some new definitions here, and made another step towards my goal. In the next post we are going to use some of the definitions here in order to construct The fundemental group(s), which will allow us to finally start involving some algebra to our theory.

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