In the last post I talked about **Categories**, and introduced maps between them (**Functors**). In this post I want to start talking about topology, and define the term of **Homotopy**.

Let be two topological spaces and continuous maps. A **Homotopy** from to is a continuous map:

H:X\times I\to Y,\ \ \ (I=[0,1])

such that for every : and . It might look a little wierd at first, but this map actually has a pretty nice intuition.

Think of the unit interval as a time parameter. We are ‘starting’ at time 0 with the function . As time goes by, we are continuously ‘transforming’ the function to the function .

I feel like this definition gets clearer when seeing a visual example. I am going to show a homotopy between two functions .

the first one, is going to ‘do nothing’ to the interval:

f(x)=(x,0)

On the other hand, is going to transform the interval to the unit circle:

g(x)=(\cos x, \sin x)

Now, our goal is to find a homtopy between them, which is not that complicated:

H(x,t)=(1-t)\cdot f(x) + t\cdot g(x) = ((1-t)x + t\cos x, t\sin x)

It’s easy to see that and . and this map is indeed continuous. Now I want you to imagine how the image of changes as increases. It will look like the straight line will become the circle. If you are struggling with imaging such a thing, don’t worry, I made an animation of it that I am sure it will make things clearer (as it did to me).

I think that’s a pretty cool animation, it helped me a lot while I was trying to understand what this definition means, and shows homotopy in a smooth, visual way.

I am going to use the notation for every , such that . With this notation, I can write . Also, If there is an homotopy between we say that ** is homotopic to **, and denote it by . This defintion makes me wonder if is an equivalence relation, (and I bet that I am not the only one) well, turns out that it is. In order to prove it we need to show that is: **reflexive**, **symetric** and **transitive**.

**Reflexive**: We need to find an homotopy between to itself, well, it is not that hard, we can define . Obviously, , the only thing left to show is that is continuous. The key here is to notice that is a composition of two functions:

- – The projection on the first components () of the product space , which is continous (in a topolgy class, we prove that a projection is indeed continous, it is even an
**open map**). - – It’s continuity is given.

Now, we can write: and derive that is continuous as a compostion of continuous maps.

X\times I \overset{\rho}{\to}X\overset{f}{\to}Y

(x,t)\overset{\rho}{\mapsto}x\overset{f}{\mapsto}f(x)

**Symetry**: given and a homotopy from to , we need to find to a homotopy from to . I am going to define as:

K(x,t)=H(x,1-t)

Think of as ‘traveling back in time’. We can verify that: and . We only need to show that is continuous. Similar to before, we are going to represent as a compostion of continuous maps. The functions are:

- , where . The way it acts is . This is indeed a continous function (again, this is proved in a topology class).
- – It’s continuity is given.

We get that is a continuous function.

X\times I \overset{Id_x\times R}{\longrightarrow}X\times I \overset{H}{\to}Y

**Trasnsitive**: We now have such that . We want to show that . Let be a homotopy from to , and a homotopy from to . Our first instinct (or at least mine) in such proofs, is to take thier composition, however, we can’t do that here, so I’ll have to take a different approach. The idea is to ‘travel’ from to twice faster, and then travel from to twice faster. Formally it looks like that:

J(x,t)=\begin{cases} H(x,2t) & 0\leq t\leq\frac{1}{2}\\ K(x,2t-1) & \frac{1}{2}\leq t\leq1 \end{cases}

Indeed, and . We only need to prove that is continous. Notice that:

J|_{X\times [0,\frac {1}{2}]}= H

J|_{X\times [\frac {1}{2},1]}=K

If you can recall, there is a theorem in topology that goes like:

“If is a function such that, are both continuous where are closed sets such that Then is continuous.”

If you are not familiar with this theorem, you can either try to prove it yourself, or look for a proof for it somewhere (In the future, I am planning on writing about point-set topology, and I will include a proof for it in there).

Since are both closed sets that cover the space and are continuous, we can conclude by using the theorem, that is indeed continuous, as required.

Great! we proved that ‘‘ is indeed an equivalence relation and we can now proceed to our next result. Suppose we have 3 spaces: and we have 2 homotopic functions and another two homotopic functions .

X\overset{f\sim g}{\to}Y\overset{h\sim l}{\to}Z

and are both elements of the same equivalence class, which I’ll denote by . The same thing applies to and as well, I am going to denote their equivalence class by . A netural question to ask now is: “Is the action well defined?” In other words, is it true that : ? Let’s try to prove that, and if it turns out to be true, we can conclude an outstanding result which involves what I have discussed on the last post!

Ok, We have 2 homotopies: where , and where . We need to find some homotpy . When I first saw it, my first instinct was to define ,however, that’s non-sense, since the output of is in and ‘s input is in . That’s a good direction though, this mistake showed me that I am on the right way, and I just need to change the composition a little bit such that get’s an input that it can ‘work with’, so my second attempt is:

is a function , defined by , which is continuous since each component is continuous (again, this statement is being proved at topology class – I’ll write about it in the future).

As you may guess, is the projection on the second component of the product space . We now have:

L: X\times I\overset{(H,\rho_2)}{\longrightarrow}Y\times I\overset{K}{\longrightarrow}Z

(x,t)\overset{(H,\rho_2)}{\mapsto}(H(x,t),t)\overset{K}{\mapsto}K(H(x,t),t)

is continuous as a composition of continuous maps, and: , as we wanted, while also: . We found a homotpy between and as required!

Now we are finally ready for our major result!

## The Homotopy Category

So far, we saw that ‘‘ is an equivalence relation, and if , then the action is well defined – It is independent of the choice of element from an equivalence class.

I am going to define now a new category – **The Homotopy Category**. It’s objects are the **Topological spaces** and the morphism are the **Homotopy classes of continuous functions**. Let’s check what is the **isomorphism** in this category: Suppose that and (recall that when I am talking about morphisms and write , I don’t mean that is a function – it is just a symbol that means ) are isomorphic. Then:

[g\circ f]=[g]\circ[f]=[Id_X]

[f\circ g] = [f]\circ [g] =[Id_Y]

Thus, and (I didn’t show that are units, however, it’s not that hard to check). In other words, two equivalence classes are isomorphic, if the compostion of two functions from the classes, will be homotopic to the identity on or (it depends on the composition).

This result allows us to define a **homotopy equivalence** between topological spaces: a homotopy equivalence between two topological spaces is a pair of functions like in the above – their composition is homotopic to one of the identity functions. If such a pair exists, we will say that and are **homotopy equivalent**.

If two spaces are **homeomorphic** then there is a pair such that and , thus, homeomorphism is **stronger** than homotopy the composition is not just homoptopic to the identity, **it is the identity**. From that, we got a new criteria to test if two topological spaces are homeomorphic – if we will be able to prove that two spaces are not homotopy equivalent, then they are not homeomorphic as well.

Great! umm… but how can we determine if the spaces are homotopy equivalent or not? Since it is a weaker property than homeomorphism, proving that two spaces are homotopy equivalent is harder!

However, not everything is lost – as we will see in the future, **groups** will come to help us!

That’s it for this post, in the next post I am going to discuss some more properties of homotopy in order to understand it better. And maybe even more than that…