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Question 1.1.16 from Hatcher:

Show that there are no retractions $r : X\to A$ for the case $X = \mathbb{R}^3$, $A$ any subspace homeomorphic to $S^1$.

There's a bunch of other cases too, but I'd just like help as to how to go about this sort of question. I can clearly visualise that there is no retraction, but how do you write this?

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  • $\begingroup$ @SL2 has set you on the right path below, but I think it's worth emphasizing that the key here is that $\pi_1$ is what is known as a functor: it takes compositions to compositions and identity maps to identity maps. Every tool in algebraic topology worth thinking about draws its power first and foremost from functoriality. If $\pi_1$ were just some gadget that took in spaces and spat out groups, we wouldn't have arguments like this. Functoriality is also sometimes called naturality (although this term isn't actually well-defined, and it tends to get used in many contexts). $\endgroup$
    – Aaron Mazel-Gee
    May 3, 2012 at 0:46

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Suppose that $r\colon \mathbb R^3\to A$ were a retract. Then $ri=1_A$ where $i\colon A\to \mathbb R^3$ is the inclusion, and so we have that the composition $\pi_1(A)\overset{i_*}\to\pi_1(\mathbb R^3)\overset{r_*}\to\pi_1(A)$ is the identity map. But the middle group is 0 and the two outer groups are $\mathbb Z$... Do you see how to finish this?

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