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Wikipedia says about deformation retraction that it "...is a special case of homotopy equivalence..."

I fail to see how this is true. Say $A \subset X$ and $F$ a deformation retraction from $id_X$ to $id_A$.

If $F$ was a homotopy equivalence we would have to have $F_0 \circ F_1 \simeq id_X$ and $F_1 \circ F_0 \simeq id_A$.

The first is impossible because $F_1(X) \subset A$.

What am I missing? Thanks for your help.

Rudy the Reindeer
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    Why is the first impossible? Also, my answer here might help. – t.b. Apr 12 '11 at 09:31
  • Impossible because if $Im(id_X) \subset A$ then it's not the $id$ function. Oh. Just realised my mistake: I can still have a homotopy.... arrgh Now I'm annoyed with myself for making the same mistake for the $n$-th time. Thanks for your patience! – Rudy the Reindeer Apr 12 '11 at 09:35
  • don't be too hard on yourself :) – t.b. Apr 12 '11 at 09:40
  • Thanks for your kind words. But I feel retarded for repeating this mistake. – Rudy the Reindeer Apr 12 '11 at 09:45
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    Mistakes like this are good - they show that you're getting deep into the material and trying to understand it! These are the exact sort of silly errors and mistakes that you almost have to make in order to learn maths well. Keep at it - you're doing fine! :) – Alex Apr 12 '11 at 15:04
  • @Theo: I just read your answer here -- Wikipedia uses the word "deformation retraction" and writes $r: X \rightarrow A$ instead of $r: X \times [0,1] \rightarrow X$. It disturbs me greatly, I would also prefer if they wrote "...a continuous map $r: X \rightarrow A $ has an associated deformation retraction if...", because a deformation retraction is a homotopy but a retraction is just a continuous function... – Rudy the Reindeer Apr 13 '11 at 08:31
  • Wikipedia is right. A retraction is simply a map $r:X \to A$ such that $ri = 1_A$ where $i:A \to X$ is the inclusion. A (strong) deformation retraction is a special retraction $r:X \to A$ in that one adds the condition and $X$ can be deformed into $A$ along $r$ (in such a way that $A$ remains fixed throughout). That is to say $ir \simeq 1_{X}$ and the homotopy can be chosen such that $H(i(a),t) = i(a)$ for all $a$.. The existence of such a homotopy is required but it is not part of the data. – t.b. Apr 13 '11 at 08:45
  • But I agree insofar as the combination of words deformation retraction suggests that the deformation (i.e. the homotopy) is part of the data. However, from the homotopic perspective it is not so interesting exactly how you deform, only that you are able to do so. – t.b. Apr 13 '11 at 09:02
  • Hatcher (on p. 2) defines "deformation retraction" to mean what Wikipedia calls "strong deformation retraction". So algebraic topologists don't all consistently use the same definitions. Unless I misunderstand Hatcher which, of course, is probable. – Rudy the Reindeer Apr 13 '11 at 11:50
  • So, if I'm not confusing things again, Hatcher would call this $F$ in $2.$ here a deformation retraction and $A$ the corresponding deformation retract. – Rudy the Reindeer Apr 13 '11 at 11:55
  • That seems to be true. I've now checked in several references and the terminology isn't consistently used. Both uses (Wikipedia's and Hatcher's) appear about equally often. – t.b. Apr 13 '11 at 12:53
  • So, using Hatcher's definitions, is it correct to not call maps of the form $r: X \rightarrow A$ deformation retractions? That is, the homotopy is part of the data in Hatcher? – Rudy the Reindeer Apr 13 '11 at 13:27
  • Yes, that's right. By the way, you should include an @Theo if you want me to see your comments. – t.b. Apr 13 '11 at 13:36
  • @Theo: I have to thank you again, this discussion has been very useful to me. Especially, since I ordered Lee's "Introduction to topological manifolds" and he uses the same definition as Wikipedia and you. – Rudy the Reindeer Apr 13 '11 at 13:44
  • No problem, my pleasure, as always :) Have fun with Lee, his books are great! See you around. – t.b. Apr 13 '11 at 13:48
  • @Theo: thanks! Yes, see you around ; ) – Rudy the Reindeer Apr 13 '11 at 14:36

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Even if $Im(id_X) \subset A$, $F_0 \circ F_1 \simeq id_X$ can still be true. Note that $\simeq$ is not $=$.

Rudy the Reindeer
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