# Does homotopy equivalence of pairs $f:(X,A)\to(Y,B)$ induce the homotopy equivalence of pairs $f:(X,\bar A)\to(Y,\bar B)$?

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

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.

EDIT: This only works if the underlying space is metrizable.

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Really pretty proof. Thank you! – Chinmay Nirkhe Mar 28 '15 at 14:45
This proof does not work. The converse of the Sequence Lemma does not necessarily hold if the space is not metrizable, and we are not assuming our spaces are metrizable. So in other words, such a sequence $\{a_k\}$ may not exist. – 1234 Jul 7 at 15:37
@1234, Valid. I've added an edit. – Marcus M Jul 7 at 17:11
Won't the proof work even in the non-metrizable case if you use nets instead of sequences? – Andreas Blass Jul 7 at 17:14