I have a question regarding algebraic topology with which I was hoping someone could help me with. I've managed to show the following:

If $f,g:S^{n-1} \to X$ are homotopic maps, then $X\sqcup_fD^n$ and $X\sqcup_gD^n$ are homotopy equivalent. I showed this by showing they were both deformation retracts of the same space, $Y=X\sqcup_F(D^n\times I)$ where $F$ is my homotopy between maps $f$ and $g$.

I feel like this result could help me with the following question:

Let $f:S^{n-1} \to X$ be a map and $g:X \to Y$ be a homotopy equivalence. Show that $X\sqcup_fD^n\simeq Y\sqcup_{f\circ g}D^n$.

Could I do this similarly as with my original problem. Since if $g$ is a homotopy equivalence between $X$ and $Y$, this tells me that $X\simeq Y$, so they are deformation retracts of a larger space, say Z.

Would I then show $X\sqcup_f D^n$ and $Y\sqcup_{f\circ g}D^n$ are deformation retracts of some larger space $W=Z\sqcup_f (D^n\times I)$? If I retract $Z$ to $X$ and $(D^n\times I)$ to $(D^n\times ${$0$}$) \cup (S^{n-1}\times I)$, is it valid for me to apply my map $g$ and yield that $Y\sqcup_{f\circ g}D^n$ is a deformation retract of my larger space $W$?

Thanks in advance.

  • 1
    $\begingroup$ I'd like to point our a common terminology mistake. Spaces are "homotopy equivalent;" maps are "homotopic." $\endgroup$ Jun 8, 2013 at 17:36
  • $\begingroup$ Yes, thank you. I'll edit that now! $\endgroup$
    – Alex
    Jun 9, 2013 at 11:12
  • $\begingroup$ The following is relevant:mathoverflow.net/questions/96071/… $\endgroup$ Mar 17, 2016 at 15:01

2 Answers 2


Okay, here is my proof which took me a while to elaborate, but I think it's correct:

Note that $(D^n,S^{n-1})$ has the homotopy extension property (HEP). The salient point of my proof is a retraction of $B\times I\times I$ onto $(A\times I\times I)\ \cup\ (B\times\{0\}\times I)\ \cup\ (B\times I\times\{0\}),$ whenever $(B,A)$ has the HEP.
For $(s,t)\in I\times I$ let $s^*(s,t)=||(s,t)-d(s,t)||/||(1,1)-d(s,t)||$. Let $r$ be the retraction of $B\times I$ onto $A\times I\ \cup\ X\times\{0\}$ with coordinates $(r_x,r_s)$, and let $d:I\times I\ \longrightarrow\ I\times\{0\}\ \cup\ \{0\}\times I$ be a retraction. Now define $R:B\times I\times I\ \longrightarrow\ (A\times I\times I)\ \cup\ (B\times\{0\}\times I)\ \cup\ (B\times I\times\{0\})$, $(x,s,t)\mapsto(r_x(x,s^*),\ \ d(s,t)+r_s(x,s^*)\cdot[(1,1)-d(s,t)])$. It is not difficult to prove that $R$ is a well-defined retraction.

Two spaces are homotopy equivalent iff they are strong deformation retracts of a larger space, and this larger space can be chosen to be the mapping cylinder $M(g)=Y\cup X\times I$, where $(x,1)\sim g(x)$. Then you can glue $B\times I$ in a canonical way, by the map $f\times\text{Id}_I$. There are maps $X\times0\xleftarrow{\ \ r\ \ }M(f)\xrightarrow{g\circ p_X}Y$ where $r$ is a deformation retraction, meaning there is a $k:M(f)\times I\to M(f),\ k(-,1)=r,\ k(-,0)=\text{Id}_{M(f)}$, a homotopy from $r$ to the identity. The map $g\circ p_X$ is a retraction homotopic to the identity via the map $h$ such that $h(x,s,t)=(x,t+s-ts)$.

Let us define a map $K:(M(f)\cup_{f\times Id} B\times I)\times I\ \longrightarrow\ M(f)\cup_{f\times Id}B\times I$ whose restriction onto $B\times I\times I$ is defined in the following way: We have shown that $B\times I\times I$ retracts onto $A\times I\times I\ \cup\ B\times(\{0\}\times I\cup I\times\{0\})$ A map on the first term is given by $k\circ(f\times\text{Id}_I\times\text{Id}_I)$, and one on the second term by $(b,s,t)\mapsto(b,s)$. They coincide on the common domain, and so they induce a map $K':B\times I\times I\to M(f)\cup_{f\times Id}B\times I$. We then combine $K'$ with $k$ on the cylinder to obtain $K$. One can then check that this is a deformation retraction onto $X\times\{0\}\cup_{f\times 0}B\times\{0\}$.

What is left is to find a deformation retraction onto $Y\cup_{gf}B$. But this one can even be explicitly written down, and the formula is practically the same as for $h$.


For X and Y CW complexes, the statement that you are trying to prove is Theorem 1 of the following reference: "Topics in Combinatorial Differential Topology and Geometry" by Robin Forman: http://books.google.com/books?id=W_SPdwfPTw8C&pg=PA135&dq=Topics+in+Combinatorial+Differential+Topology+and+Geometry%22+by+Robin+Forman&hl=en&sa=X&ei=1BryU4LoCpPgsAT7hYGoAQ&ved=0CB4Q6AEwAA#v=onepage&q=Topics%20in%20Combinatorial%20Differential%20Topology%20and%20Geometry%22%20by%20Robin%20Forman&f=false.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.