# Example for not deformation retraction but retraction

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.

• What is $B^2$? I cannot really understand the example. – N.B. Oct 29 '18 at 10:56
• @N.B. Sorry. I should have posted it more detail. $B^2$ is unit ball in $R^2$ – fivestar Oct 29 '18 at 10:58
• 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 $X$? – 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.

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.

• I almosr understand your explaination . But I want to ask wht connectedness is homotopy invariance – fivestar Oct 29 '18 at 15:40
• @fivestar read this: math.stackexchange.com/questions/1755384/homotopy-connectedness – freakish Oct 29 '18 at 15:42
• Thank you! I ‘ll refer to your link! – fivestar Oct 29 '18 at 15:48
• 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) – Max Oct 29 '18 at 17:06
• @Max OP edited his question. That's not what he wrote orignally. – freakish Oct 29 '18 at 17:28