# Deformation retraction: special case of homotopy equivalence?

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.

• Why is the first impossible? Also, my answer here might help. – t.b. Apr 12 '11 at 9:31
• 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 9:35
• don't be too hard on yourself :) – t.b. Apr 12 '11 at 9:40
• Thanks for your kind words. But I feel retarded for repeating this mistake. – Rudy the Reindeer Apr 12 '11 at 9:45
• 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

## 1 Answer

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