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What is an example of topological spaces $X \subseteq Y$ such that $X$ is closed in $Y$, and $X$ and $Y$ are both contractible, yet $Y$ does not retract to $X$? I'm having a hard time coming up with a quick example. Requiring $X$ to be closed in $Y$ seems to rule out the case where $[0, 1]$ does not retract to $(0, 1)$. Or is there no such example, and does such $Y$ always retract to such $X$?

Vikram Saraph
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  • You can find examples in the exercises of hatcher chapter 0 of contractible spaces that don't deformation retract to any point. For one of them (the infinite comb space, or whatever he calls it), I would bet it doesn't deformation retract onto any nontrivial closed subset. –  Dec 01 '15 at 19:05
  • Yeah, I saw the example of the comb space. The conclusion is that the comb space doesn't deformation retract to any point, and I'm willing to buy that that generalizes to any closed subspace, but would that necessarily imply the nonexistence of a plain old retraction? – Vikram Saraph Dec 01 '15 at 19:08
  • Oh, sorry, I completely misread your question. I suspect the comb doesn't retract onto the zig-zag, but haven't checked. –  Dec 01 '15 at 19:09
  • Actually, I think the comb does retract onto the zig-zag, by projecting each bristle straight up or down. – MartianInvader Dec 01 '15 at 22:31

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Let $Y = \mathbb{R}^2$.

For each positive integer $n$, let $A_n$ be the segment $\{\frac{1}{n}\} \times [0,1] \subset \mathbb{R}^2$. Let

$$ X = \left(\{0\}\times [0,1]\right) \cup \left([0,1] \times \{0\}\right) \cup \bigcup_{i}A_i $$

Then $X$ is closed in $Y$ and is contractible. But there is no retraction $\mathbb{R}^2 \rightarrow X$.

To see this, suppose $f\colon \mathbb{R}^2\rightarrow X$ is a retraction, and consider the segment $S = [0,1] \times \{1\}\subset \mathbb{R}^2$. This segment intersects $X$ at all the points $\left(\frac{1}{n}, 1\right)$, so the sub-segments between these points must be mapped by $f$ to paths along $X$ connecting them. In particular, each such sub-segment must have a point $s_n$ mapped to $\left(\frac{1}{n}, 0\right)$, since any path from $\left(\frac{1}{n}, 1\right)$ to $\left(\frac{1}{n+1}, 1\right)$ in $X$ passes through this point.

Now consider the sequence $f(s_n)$. On the one hand, it must converge to $\left(0,1\right)$ since the $s_n$ converge to $\left(0,1\right)$, which is fixed by $f$ since it lies in $X$. On the other hand, since $f(s_n) = \left(\frac{1}{n}, 0\right)$, the sequence must converge to the origin, giving a contradiction.

MartianInvader
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