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A deformation retraction in the weak sense of a space $X$ to a subspace $A$ is a homotopy $f_t:X\to X$ such that $f_0=Id_X$, $f_1(X)\subset A$ and $f_t(A)\subset A$ for all $t$. Show that if $X$ deformation retracts to $A$ in this weak sense, then the inclusion $i:A\to X$ is a homotopy equivalence.

Proof: We need to find a $g:X\to A$ such that $i\circ g=Id_X$ and $g\circ i=Id_A$. If we put $g=f_1$, we see that $i\circ f_1=f_1\simeq Id_X$, exactly because of the deformation retraction. Now, as $f_1\simeq Id_X$, $f_1\circ i\simeq Id_X\circ i=Id_A$.

Are there any mistakes?

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  • $\begingroup$ I don't see any mistake. $\endgroup$ – Stefan Hamcke Jul 4 '13 at 15:12
  • $\begingroup$ You are spot on. $\endgroup$ – Vishal Gupta Jul 4 '13 at 15:22
  • $\begingroup$ There's a slight issue at the end, I think. You want to have a map $A \to A$ that homotopes to the identity, and what you've given is a map $A \to X$. So maybe we should be more careful about cutting down $f_t$ appropriately and explicitly use the condition $f_t(A) \subset A$. $\endgroup$ – TTS Jul 4 '13 at 15:30
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This answer is just to remove the question from the unanswered queue.

You are right in what you have done. Could you be more specific about your doubt? Which step are you unsure of?

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  • $\begingroup$ I was actually unsure of the very last step (sometimes I am confused about basic set theoretical rules) $\endgroup$ – Xena Jul 4 '13 at 15:27
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I agree with the comment by TTS, you need to be a bit more careful in the last step. Your argument does not make use of the fact that $f_t(A)\subset A$ for all $t$, which is crucial.

You are right that $f_1\simeq Id_X$ as maps $X\to X$ but this does not immediately imply that $f_1\circ i$ and $Id_A$ are homotopic as maps $A\to A$. What you need is that $f_1|_A\simeq Id_X|_A$ as maps $A\to A$. To show this the assumption $f_t(A)\subset A$ for all $t$ is required.

To illustrate what might go wrong if we dropped the assumption $f_t(A)\subset A$ for all $t$, consider the following example.
Let $S^n\subset \mathbb{R}^{n+1}$ be the unit sphere of odd dimension $n>1$. Then the antipodal map $$ -Id:S^n\to S^n: x\mapsto -x $$ is homotopic to the identity map $Id_{S^n}$ via $$ f_t(x) = \cos(t) x+\sin(t)v(x) \quad t\in[0,\pi]\quad f_0=Id_{S^n} \quad f_\pi=-Id $$ where $x=(x_1,\ldots,x_{n+1})\in S^n$ and $v(x)=(x_2,-x_1,x_4,-x_3,\ldots,x_{n+1},-x_n)\in S^n$.
Now we intersect $S^n$ with any $3$-dimensional linear subspace of $\mathbb{R}^{n+1}$ in order to get a $2$-sphere $S^2\subset S^n$. Even though $-Id$ and $Id_{S^n}$ are homotopic as maps $S^n\to S^n$, their restrictions $$-Id|_{S^2}\text{ and }Id_{S^n}|_{S^2}$$ to $S^2\subset S^n$ are not homotopic as maps $S^2\to S^2$. The reason for this is because they don't have the same degree ($1=\deg(Id|_{S^2})\neq\deg(-Id|_{S^2})=(-1)^3=-1$, see Section 2.2 in Hatcher's book) and the degree of a smooth map is invariant under homotopy.
Notice that, whatever $2$-sphere $S^2\subset S^n$ you choose, the homotopy always violates the condition:
$$f_t(S^2)\subset S^2 \text{ for all }t\in[0,\pi].$$

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