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My professor introduce following example to explain the difference of retraction and deformation retraction.

Consider $X=B^2$. Let $A=${lower part of $B^2$}

I can find $A$ is retract of $X$, by giving projection map on upper part and identity map on lower part. Then using pasting lemma guarantee the continuity.

But, I don't know how to show $X$ doesn't have deformation retract. Please explain why the space doesn't have deformation retract. Thank you.

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    $\begingroup$ What is $B^2$? I cannot really understand the example. $\endgroup$ – N.B. Oct 29 '18 at 10:56
  • $\begingroup$ @N.B. Sorry. I should have posted it more detail. $B^2$ is unit ball in $R^2$ $\endgroup$ – fivestar Oct 29 '18 at 10:58
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    $\begingroup$ From Wikipedia: " a retraction is a continuous mapping from a topological space into a subspace..." [emphasis mine]. Note that $A$ is not a subset of $X$ (indeed, it's a piece of $B^2$ that's been removed from $X$!), so it's hard to see how it could be a retract of $X$. Are you sure you've got the right definition for $X$? $\endgroup$ – John Hughes Oct 29 '18 at 11:48
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In order for $A$ to be a deformation retract of $X$, $A$ has to be a subset of $X$. Thus your example makes little sense. But we can fix it.

Let

$$X=\big\{(x,y)\in\mathbb{R}^2\ \big|\ \lVert(x,y)\rVert\leq 1\text{ and }y\neq 0\big\}$$ $$A=\big\{(x,y)\in\mathbb{R}^2\ \big|\ \lVert(x,y)\rVert\leq 1\text{ and }y> 0\big\}$$

This time $A\subseteq X$ and there's an obvious retraction

$$r:X\to A$$ $$r(x,y)= (x,|y|)$$

But $A$ is not a deformation retract of $X$. Indeed, $X$ cannot be homotopy equivalent to $A$ because $X$ has two connected components while $A$ has one (in other words: connectedness is a homotopy invariant). And a deformation retraction is a special case of homotopy equivalence.

This example can be simplified even further. Let $X=\{-1,1\}$ (with discrete topology) and $A=\{1\}$. Then $A$ is a retract of $X$ (via constant function $x\mapsto 1$) but not a deformation retract. You can't get it simplier than that. :D


For a more sophisticated example (i.e. a one that doesn't involve connectedness argument) consider the sphere $S^1=\{v\in\mathbb{R}^2\ |\ \lVert v\rVert=1\}$ and a point $x\in S^1$. Then $\{x\}$ is an obvious retract of $S^1$ (a singleton is a retract in any space) but $\{x\}$ is not a deformation retract since $S^1$ is not contractible. The choice of $S^1$ is not really necessary, any non-contractible space will do.

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  • $\begingroup$ I almosr understand your explaination . But I want to ask wht connectedness is homotopy invariance $\endgroup$ – fivestar Oct 29 '18 at 15:40
  • $\begingroup$ @fivestar read this: math.stackexchange.com/questions/1755384/homotopy-connectedness $\endgroup$ – freakish Oct 29 '18 at 15:42
  • $\begingroup$ Thank you! I ‘ll refer to your link! $\endgroup$ – fivestar Oct 29 '18 at 15:48
  • $\begingroup$ I don't see what's wrong with the example in the OP, with $X= B^2$, $A=B^2\cap (\mathbb{R}\times\mathbb{R}_{-})$ , we do have $A\subset X^2$ (of course with $B^2$ this is a deformation retract, but with $S^1$ for instance it would be an interesting example) $\endgroup$ – Max Oct 29 '18 at 17:06
  • $\begingroup$ @Max OP edited his question. That's not what he wrote orignally. $\endgroup$ – freakish Oct 29 '18 at 17:28

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