11
$\begingroup$

Okay, I keep reading the definitions over and over, but I don't see the difference between the two ; apparently spaces that deformation retract onto a point are contractible, but the opposite is not necessarily true.

Definition of contractible : A space $X$ is said to be contractible if it has the homotopy-type of a point, i.e. there is a map $f : X \to \{p\}$ and a map $g : \{p\} \to X$ such that $fg \simeq \mathbb 1_p$ and $gf \simeq \mathbb 1_X$. Since $fg = \mathbb 1_p$, it suffices to show that there is some point in $X$ such that the identity map $\mathbb 1_X$ is homotopic to a constant map at some point $g(p) \in X$.

Definition of deformation retract onto a point : there exists a map $f : X \times I \to X$ such that $f(x,0) = \mathbb 1_X$ and $f(x,1) = p$ for some fixed point $p \in X$ and $p$ does not depend on $x$.

I believe an homotopy from the identity map to the constant map is precisely a deformation retract, so I don't see the difference. Or maybe I got the definitions wrong. Does anybody see it? I am quite stuck, since one exercise asks to show that some weird space is contractible but does not deformation retract onto a point.

$\endgroup$
2
10
$\begingroup$

For a contractible space, the homotopy is allowed to move the point, we don't need to have

$$h(p,t) = p \text{ for all } t \in [0,1].$$

For the (strong(1)) deformation retract, $p$ must be kept fixed by the homotopy.

So (strong) deformation-retractability is a stronger condition.

(1) Different authors use different terminology. Some call simply deformation retract what others call strong deformation retract, and that seems to be the case if "spaces that deformation retract onto a point are contractible, but the opposite is not necessarily true". If a deformation retract need not keep the retract fixed in the homotopy, the two properties are exactly the same.

$\endgroup$
4
  • $\begingroup$ Ooooooh, right. Did not see that difference. Thanks for pointing it out! $\endgroup$ – Patrick Da Silva Sep 1 '13 at 14:19
  • 2
    $\begingroup$ Yes, @user8268 beat me to the punch. Different authors have different definitions. I grew up with the definition that allows Patrick to be confused. Daniel's definition is, for me, a strong deformation retraction. So, Patrick, check your book's definition carefully! $\endgroup$ – Ted Shifrin Sep 1 '13 at 14:23
  • $\begingroup$ As a simple example, a torus ($S^1\times S^1$) surely cannot deformation retract to a point, but of course is contractible. $\endgroup$ – dezign Sep 1 '13 at 15:10
  • 16
    $\begingroup$ @dezign A torus isn't contractible. $\pi_1(\mathbb{S}^1\times \mathbb{S}^1) \cong \mathbb{Z}^2$. $\endgroup$ – Daniel Fischer♦ Sep 1 '13 at 15:25

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.