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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.

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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.

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

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