Why are retracts important?

Question

This post was helpful in understanding what retracts are but I am still confused on why one would wish to study them. I think it would be helpful if it could be said how the concept originally emerged.

Answer 1

Here is why I think retracts are useful: consider an injective map $f : X\to Y$ between spaces. Since this map is injective, it's a natural question to ask if the induced map $f_* : \pi_1(X) \to \pi_1(Y)$ is also injective. This seems reasonable, but in fact it's not true! Consider for instance the embedding $i : S^1 \hookrightarrow D^2$. This map is injective but clearly $f_* : \pi_1(S^1)\cong\mathbb{Z} \to 0\cong\pi_1(D^2)$ is not.

The reason that this didn't work is that even though $i$ is injective, it does not have a continuous left-inverse $r : D^2 \to S^1$, since $S^1$ is not a retract of $D^2$ (intuitively, you would have to punch a hole through the $D^2$, which destroys continuity).

However, if we have a retract $r : X \to A$ for some injection $i : A \hookrightarrow X$, then this means in particular that $r$ is continuous, i.e. a morphism in the category of topological spaces. This will allow us to use functoriality of $\pi_1$, namely we see that the composition $$ \pi_1(A) \xrightarrow{i_*} \pi_1(X) \xrightarrow{r_*} \pi_1(A) $$ is the same as $\text{id}_{\pi_1(A)}$, since $$ r_* \circ i_* = (r \circ i)_* = (\text{id}_A)_* = \text{id}_{\pi_1(A)}. $$ Thus we have shown that the homomorphism $i_*$ has a left-inverse, i.e. $i_*$ is indeed injective now. Of course the same result holds for any other covariant functor, such as homology $H_n(-; \mathbb{Z})$, which will be useful later. Do you see what happens when we use a contravariant functor?

In my opinion this is a neat result, because it's always nice when properties translate when you jump between different categories. I hope this was helpful!

Answer 2

Here is just one answer: it is more algebraic, from the perspective of "we care about the simplest things the most," rather than "we care about things that give interesting geometric data."

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jasnee is correct in that retracts have a special role in algebraic topology, but that special role is actually a symptom of a more general, categorical reason.

A helpful fact of set theory that most students learn is that a function $f$ is injective if and only if it has a left inverse; i.e., there is some other function $g$ so that $gf = \text{id}$. We can pull this definition back to a categorical notion: we define a retract to be a left inverse of some morphism $m$. It is short but worthwhile to prove that in $\mathsf{Set}$, a morphism is given by an injective map if and only if it has a retract. There are very few simpler structures in a category than inverses of morphisms, so categorical retracts are in themselves a fundamental notion.

One reason why we care about topological retracts is because it is a "geometrization" of this fundamental notion. Indeed, if we work in $\mathsf{Top}$, then morphisms carry with them geometric meaning, as continuous maps induce topologies. The crux of this discussion is the following:

A topological retract $X \to A$ is the categorical retract of the inclusion map $A \hookrightarrow X$.

Hence this geometric notion of "deforming a space to a subspace" is really just a topological interpretation of a fundamental notion, so it's bound to crop up as a point of study.

This also explains why we like retracts for algebraic topology; because of the functoriality of $\pi_1$, it follows that a topological retract induces a group-theoretic retract of the group homomorphism $\pi_1(A) \hookrightarrow \pi_1(X)$.

Answer 3

One way to learn the usefulness of a mathematical concept is to see how it is used (I suppose that is a tautology?).

Here is perhaps one of the most striking application of retracts that a first year algebraic topology student learns.

Brouwer fixed point theorem: Every continuous function $f : B^n \to B^n$ has a fixed point.

Outline of the proof:

• Step 1: If $f$ does not have a fixed point then one may use $f$ to construct a topological retract $r : B^n \to S^{n-1}=\partial B_n$.
• Step 2: By applying the $n^{\text{th}}$ homology functor one obtains an abelian groups retract $r_* : H_{n-1}(B^n) \to H_{n-1}(S^{n-1})$.
• Step 3: Since $H_{n-1}(B^n) \approx 0$ and $H_{n-1}(S^{n-1}) \approx \mathbb Z$, one obtains a retract $0 \to \mathbb Z$. This is a contradiction.

Step 2 is an example of a very useful principle of category theory: every functor takes retracts of the domain category to retracts of the range category.