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?
Does homotopy equivalence of pairs $f:(X,A)\to(Y,B)$ induce the homotopy equivalence of pairs $f:(X,\bar A)\to(Y,\bar B)$?
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0It turned out that we could use the property that f is continuous and a limit can go in and out of those continuous ftns. – 2012-04-04
2 Answers
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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0Really pretty proof. Thank you! – 2015-03-28
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0This 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. – 2016-07-07
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0@1234, Valid. I've added an edit. – 2016-07-07
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0Won't the proof work even in the non-metrizable case if you use nets instead of sequences? – 2016-07-07
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 $S$.