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Let $h \colon A \to B$ and $r \colon S^{n-1} \to A$ be continuous maps. Assume that $h$ is an homotopy equivalence, prove that $$ D^n \cup_{r} A \simeq D^n \cup_{h \circ r} B$$

where $D^n \cup_{r} A$ is the push out of the square

enter image description here

And the second is analogue.

My attempt

Being $h$ an homotopy equivalence implies the existence of a map $s \colon B \to A$ such that $A\circ B \simeq Id$ $B\circ A \simeq Id$.

We already proved that if $a \simeq b$ maps from $S^{n-1} \to X$ then $X \cup_f D^n \simeq X \cup_g D^n$. So we have, with the previous notation, $$D^n \cup_{r} A \simeq D^n \cup_{s\circ h \circ r} A $$ Moreover $h$ induces a map $$\bar{h} \colon D^n \cup_{r} A \to D^n \cup_{h \circ r} B$$ ("extend" $h$ on $D^n$ to be the identity - so we are considering a map from the disjoint union- and then is easily verified that factors through the quotient), in the same fashion define $\bar{s}$. So the composition $\bar{h}\circ \bar{s}$ is a map $$ D^n \cup_{r} A \to D^n \cup_{h \circ r} B \to D^n \cup_{s \circ h \circ r} A \simeq D^n \cup_{r} A $$ But I cannot prove that the homotopy equivalence factors through the quotient, because it must induce another homotopy equivalence and I don't feel I have much control over the last homotopy equivalence. Some hints?

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  • $\begingroup$ possible duplicate of Homotopy equivalence of two different gluings of $B^n$ and an arbitrary space $X$ $\endgroup$ – Zhen Lin Nov 22 '14 at 9:11
  • $\begingroup$ @ZhenLin mmh it's not so clear why the question you linked should provide an answer to me. I've said that I've proved the result you linked, and my problem is that I don't know how to use it to obtain my result which is slight different according to me. Maybe I don't see some obvious properties, (this is way I asked here for clarifications) but otherwise I don't think it's a duplicate. $\endgroup$ – Luigi M Nov 22 '14 at 9:23
  • $\begingroup$ Sorry, I mean this one instead. $\endgroup$ – Zhen Lin Nov 22 '14 at 9:31

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