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When we have a homotopy equivalence through a pair $f:(X,A)\to (Y, B) $, it is said that we can induce a homotopy equivalence through a pair $f:(X,\bar A)\to (Y,\bar B) $, where $\bar A$ stands for the closure of A. Do you know how we can prove this?

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It turned out that we could use the property that f is continuous and a limit can go in and out of those continuous ftns. –  Emily Apr 4 '12 at 15:36

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I came across this post and have just solved this problem for a course I am taking, so I thought I'd post my solution; as you said, it pretty much boils down to interchanging limits via continuity:


If $a \in \overline{A}$, then $\exists~\{a_k\} \subset A$ s.t. $\lim\limits_{k \to \infty} a_k = a$. Then we have \begin{equation*} f(a) = f(\lim\limits_{k \to \infty} a_k) = \lim\limits_{k \to \infty} f(a_k) \in \overline{B} \end{equation*} where limits are interchanged due to continuity of $f$, and last statement is due to $f(a_k) \in B$ for all $k$. We now have $f(\overline{A}) \subset \overline{B}$. $f$ homotopy equivalence of pairs $\implies \exists~g:Y \to X$ and $F_t:X \to Y$ with. $F_0 = g \circ f, F_1 = \text{id}_X$ and $F_t(A) \subset A$ for all $t$. For $a \in \overline{A}$ as above, we have \begin{equation*} F_t(a) = F_t(\lim\limits_{k \to \infty } a_k) = \lim\limits_{k \to \infty } F_t(a_k) \in \overline{A} \end{equation*} Thus $F_t(\overline{A}) \subset \overline{A}$ for all $t$. This implies that $f:(X,\overline{A}) \to (Y,\overline{B})$ is a homotopy equivalence of pairs.

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Really pretty proof. Thank you! –  Chinmay Nirkhe Mar 28 at 14:45

It is not true. For example, let $$X=[0,3],\quad A=[0,1)\cup (2,3],\\ Y=[0,2],\quad B=[0,1)\cup(1,2]$$ Then the map $f: X\to Y$ which sends $[0,1]$ to $[0,1]$, $[1,2]$ to $\{1\}$, and $[2,3]$ to $[1,2]$ is a homotopy equivalence since it shrinks to a point the contractible subspace $X\setminus A$, and $(X,X\setminus A)$ is cofibered, or you could just argue that both $X$ and $Y$ are contractible. The map $f$ restricts to a homeomorphism $A\to B$. However, $\overline A=[0,1]\cup[2,3]$ and $\overline B= Y$.

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