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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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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.

As the previous posted solution by Marcus M only holds if $X$ and $Y$ are metric spaces, I will provide a more general answer.

Assume $f:(X,A) \to (Y,B)$ is a homotopy equivalence of pairs.

Show that $f:(X,\overline A) \to (Y,\overline B)$ is a

homotopy equivalence of pairs, i.e. there exists a $g:(Y,\overline B)

\to (X,\overline A)$ such that $g \circ f \simeq I_X$ and $f \circ g

\simeq I_Y$ are homotopies of pairs.

First, we show that $f:(X,\overline A) \to (Y,\overline B)$ is well defined map of pairs, i.e. $f(\overline A) \subseteq \overline B$.

As $f$ is continuous, we know that $f(\overline A) \subseteq \overline{f(A)}$. (A proof for this will be provided in the end.) As also per assumption $f(A) \subseteq B$, so $\overline {f(A)}\subseteq \overline B$. Hence in summary $f(\overline A) \subseteq \overline B$.

Now we show that $f:(X,\overline A) \to (Y,\overline B)$ is a homotopy equivalence pairs.

Per assumption, we know that there is a continuous map $H: X \times I \to X$ such that $H(0,\cdot)=g\circ f$, $H(1,\cdot)=I_X$ and $H(A\times I)\subseteq A$. This homotopy is also homotopy of the same maps, but defined for the pair $(X,\overline A)$, i.e. $H(\overline A \times I)\subseteq \overline A$ (which we need to check).

But this is exactly the same situation as in the proof of $f(\overline A) \subseteq \overline B$ if we replace $f$ with $H$, $A$ with $A \times I$ and $B$ with $A$.

The proof for the second homotopy of pairs follow analogously.

Lemma:Let $f:X\to Y$ be continuous.

Then for any subset $S\subseteq X$, it holds that $f(\overline S)\subseteq \overline{f(S)}$.

For contradiction, assume $f(\overline S)\not \subseteq \overline{f(S)}$, i.e. there exists $x \in \overline S$ such that $f(x) \in Y\setminus \overline{f(S)}$, which is open. Hence there exists an open neighborhood $U\subseteq Y\setminus \overline{f(S)}$ of $f(x)$. As $f$ is continuous this implies that $f^{-1}(U)\subseteq X\setminus \overline S$ is an open neighborhood of $x$. Further, for any $z \in f^{-1}(U)$, $f(z)\in U\subseteq Y\setminus \overline{f(S)}\subseteq Y\setminus f(S)$, thus $z \not \in S$. Therefore, $f^{-1}(U)$ is an open neighborhood of $x$ which does not intersect $S$. But then $x$ cannot be in the closusre of $S$.

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