In general, how do you prove that a topological space isn’t a retraction of another?

Question

For instance, let’s say I want to prove that $S^1$ isn’t a retract of $\mathbb{R}^3$. One way I thought of doing this was noticing that the fundamental group of $S^1$ isn’t a subgroup of the fundamental group of $R^3$. Is that correct? Is there another way to do it?

Answer 1

A space $X$ has the FPP (fixed point property) if every continuous $f: X \to X$ has a point $x \in X$ so that $f(x)=x$.

1. If $Y$ is a retract of $X$ and $X$ has the FPP, so then $Y$ also has the FPP (quite easy to prove yourself from the definitions).
2. $S^1$ does not have the FPP (consider a non-trivial rotation, which has no fixed point).
3. $B^2$ or in general any $B^n$ has the FPP (Brouwer's fixed point theorem, a non-trivial result, which has a purely combinatorial proof, among others).

So in your case of interest, if we have $S^1 \subseteq \Bbb R^3$ (via some embedding), suppose a retraction existed $r: \Bbb R^3 \to S^1$. Then expand $S^1$ as the boundary circle of some copy of $B^2 \subseteq \Bbb R^3$ and note that the restricting of $r$ to $B^2$ is still a retraction onto $S^1$, but then 1-3 imply an immediate contradiction.

So no retraction exists.

Answer 2

Homology is a nice way of doing it. If $A$ is a retract of $X$ then we have a short exact sequence $0\rightarrow H_n(A)\rightarrow H_n(X)\rightarrow H_n(X,A)\rightarrow 0$ that splits. Hence $H_n(X)\cong H_n(X,A)\oplus H_n(A)$

Noticing that $\mathbb{R}^2$ is convex and thus contractible, we may use the isomorphism above to obtain a contradiction in the example you mentioned.