Let $(X,A)$ and $(Y,B)$ be pairs of spaces and subspaces, let $\operatorname{Map}(X,Y)$ the space of maps $f:X\to Y$ equipped with the compact-open topology and let $\operatorname{Map}(X,A;Y,B)$ be the subspace of maps $f:X\to Y$ such that $f(A)\subseteq B$. Suppose the inclusions $A\hookrightarrow X$ and $B\hookrightarrow Y$ are cofibrations, would that be enough to ensure the inclusion $\operatorname{Map}(X,A;Y,B)\hookrightarrow\operatorname{Map}(X,Y)$ is a cofibration or are other conditions needed?

In particular if $X$ and $Y$ are well-pointed is the inclusion of the based mapping space $\operatorname{Map}_*(X,Y)\hookrightarrow\operatorname{Map}(X,Y)$ a cofibration?

It seems like this should be true for reasonably nice spaces and there are similar results. I know, for example, that if $B\hookrightarrow Y$ is a closed cofibration and $X$ is compact Hausdorff then the inclusion $\operatorname{Map}(X,B)\hookrightarrow\operatorname{Map}(X,A;Y,B)$ is a cofibration. In particular this makes based mapping spaces with compact Hausdorff domain well-pointed.


1 Answer 1


The inclusion $\operatorname{Map}(X,A;Y,B)\hookrightarrow \operatorname{Map}(X;Y)$ will be a cofibration whenever $(X,A)$ and $(Y,B)$ are cofibrations, under some mild point-set hypotheses. My argument requires $Y$ to be Hausdorff and $A$ and $X$ to be locally compact Hausdorff (but I'm not sure if these restrictions are necessary).

To begin with, recall that the map given by restriction $p\;\colon \operatorname{Map}(X;Y)\to \operatorname{Map}(A;Y)$ is a fibration when $A$ and $X$ are locally compact Hausdorff. This is Theorem 2.8.2 of Spanier's book.

Now it is a result of A. Strøm (Math. Scand. 22 130–142 (1969)) that if $p\;\colon E\to B$ is a fibration and $i\;\colon C\hookrightarrow B$ is a closed cofibration, then $p^{-1}(C)\hookrightarrow E$ is a cofibration.

To complete the argument, therefore, it suffices to show that $i\;\colon \operatorname{Map}(A;B)\hookrightarrow\operatorname{Map}(A;Y)$ is a closed cofibration. To see this, use the characterization of cofibrations in terms of retractions onto mapping cylinders. We have a retraction $r_Y\;\colon Y\times I \to Y\times\lbrace 0\rbrace \cup B\times I$. We can use this to define a retraction $r\;\colon\operatorname{Map}(A;Y)\times I \to \operatorname{Map}(A;Y)\times \lbrace 0\rbrace \cup \operatorname{Map}(A;B)\times I$ by setting $$ (f,t)\mapsto \big( (a\mapsto p_Yr_Y(f(a),t)),t \big), $$ where $p_Y$ is the projections from the mapping cylinder onto $Y$. This shows $i$ is a cofibration; it is closed since $\operatorname{Map}(A;Y)$ is Hausdorff.

  • 1
    $\begingroup$ The second entry of your formula for r has a free occurance of a. I guess it should be $\inf_{a \in A} p_I r_Y (f(a),t)$. $\endgroup$ Aug 31, 2012 at 19:16
  • $\begingroup$ @Karol: You are right. I have edited it to something simpler, which I think works. $\endgroup$
    – Mark Grant
    Aug 31, 2012 at 19:48
  • $\begingroup$ Thanks Mark, that's exactly the kind of argument I've been looking for! $\endgroup$ Sep 1, 2012 at 12:23
  • $\begingroup$ @Richard: You're welcome, I had fun thinking about it! $\endgroup$
    – Mark Grant
    Sep 3, 2012 at 9:35

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.