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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.