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Let $X$ be a topological space and $A\subset B\subset X$. If $A$ and $B$ are deformation retracts of $X$, then is it necessarily true that $A$ is a deformation retract of $B$? I see that at least the inclusion $A\to B$ must be a homotopy equivalence, but I can't see whether $A$ is a deformation retract of $B$.

blancket
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  • See here https://math.stackexchange.com/questions/37736/x-and-y-are-homotopy-equivalent-leftrightarrow-exists-z-x-y-are-str – Sumanta Sep 21 '20 at 11:54

1 Answers1

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Let $(i_B:B\to X,r_B : X\to B,H_B:1_X\simeq i_Br_B)$ be the data of the deformation retraction for $B$, $(i_A,r_A,H_A)$ the corresponding data for $A$.

Let $i:A\to B$ be the inclusion, so $i_A=i_Bi$.

Then, first it's clear that $A$ is a retract of $B$, since if $r_A:X\to A$ is the retraction, $r:=r_Ai_B : B\to A$ satisfies $ri=r_Ai_Bi = r_Ai_A=1_A$. The question then boils down to whether or not we can find a homotopy $H:1_B\simeq ir$.

Well, for this we have we have $$i_Br_B\simeq 1_X\simeq i_Ar_A = i_Bir_A.$$ Now postcompose with $r_B$ and precompose with $i_B$ to get a homotopy $$1_B\simeq r_Bi_Bir_Ai_B = ir,$$ as desired.

Note that if these are strong deformation retractions, in that the homotopies are the identity on the subspace, then the homotopy we produced should also be the identity on $A$.

jgon
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