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Let $(X,A)$ be a CW-pair, $Y$ a CW-complex, and $f,g:X\to Y$ homotopic maps such that $f_{|A}=g_{|A}$. Even though $f$ and $g$ are homotopic, they do not have to be homotopic relative $A$. (Obstruction theory tells us how to deal with this issue.)

Let us further assume that $B\subseteq Y$ is a contractible subcomplex and that the compositions $X\stackrel{f}{\longrightarrow}Y\stackrel{pr}{\longrightarrow} Y/B$ and $X\stackrel{g}{\longrightarrow}Y\stackrel{pr}{\longrightarrow} Y/B$ are homotopic relative $A$.

I haven't thought about this for very long, but shouldn't this already imply that the original maps $f$ and $g$ are homotopic relative $A$?

(If this is true, then we can probably drop the assumption of $f$ and $g$ being (freely) homotopic.)

Sebastian

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I think that this might be true if you require that B is contractible and that its inclusion map is a cofibration. –  Harry Gindi Feb 23 '11 at 20:44
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Well, since $B$ is assumed to be a subcomplex of the CW-complex $Y$, the inclusion $B\hookrightarrow Y$ is automatically a cofibration. –  Sebastian Feb 23 '11 at 21:03
    
Sure, but it's not necessarily an $A$-cofibration. –  Harry Gindi Feb 23 '11 at 21:43

3 Answers 3

up vote 6 down vote accepted

Sure. The fact that $A$ is a subcomplex of $X$ implies that the restriction maps $$Map(X,Y)\to Map(A,Y)$$ and $$Map(X,Y/B)\to Map(A,Y/B)$$ are Serre fibrations. The fact that $B$ is a subcomplex of $Y$ and contractible implies that the projection $Y\to Y/B$ is a homotopy equivalence, which in turn implies that the resulting maps $$Map(X,Y)\to Map(X,Y/B)$$ and $$Map(A,Y)\to Map(A,Y/B)$$ are homotopy equivalences and in particular weak equivalences. It follows that the associated map of fibers $$Map(X,Y \ rel A)\to Map(X,Y/B\ rel A)$$ is also a weak equivalence, in particular injective on $\pi_0$.

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Ah! I should have seen that. –  John Klein Feb 24 '11 at 19:48
    
Great! Thank you very much! –  Sebastian Feb 24 '11 at 22:07

This example is wrong (sorry!):

Try $(X,A) = (D^n, S^{n-1})$, $Y = D^n$, $f_A = \mathrm{in}_{S^{n-1}}$, and $B = Y$. Then we have plenty of nonequivalent (rel. $A$) maps $f, g: X\to Y$, and they all become equal in $Y/B = *$.

Since $D^n$ is convex, any two maps $X\to D^n$ are homotopic by straight-line homotopies; and if the maps agree on some subset of $X$, the homotopy will be constant on that subset.

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Wait, I am confused and I should really know this. But according to Theorem 4 in Mosher and Tangora, if $K$ ($=D^n$ in our case) is a CW-complex with n-skeleton $K^n$ ($=D^n$) and if $f,g:K\to Y$ are maps which agree on $K^{n-1}$ ($=S^{n-1}$), the restrictions of $f$ and $g$ to $K^{n}$ are homotopic relative $K^{n-1}$ if the so-called difference cochain $d(f,g)\in C^n(K;\pi_n(Y))$ vanishes. As in our case $Y=D^n$ and $\pi_n(D^n)=0$, we have $d(f,g)=0$ for all $f,g:D^n\to D^n$ which agree on $S^{n-1}$ and thus all such maps $f,g:D^n\to D^n$ should be homotopic relative $S^{n-1}$!? –  Sebastian Feb 23 '11 at 22:50
    
Since you're asking for homotopies relative to $S^{n-1}$, the difference cochain probably lies in $C^n (K, \pi_n(D^n,S^{n-1}))$. This is a great example of formalism making things unclear. –  Jeff Strom Feb 23 '11 at 23:45
    
@Jeff: in your example the space of maps $D^n \to D^n$ which are fixed on $S^{n-1}$ is the same thing as the fiber at the inclusion of the fibration $F(D^n,D^n) \to F(S^{n-1},D^n)$ given by the the restriction map of function spaces. This fibration is a map from one contractible space to another, so the fiber is contractible. In other words, I dont see how one gets non-homotopic maps $D^n \to D^n$ relative to $S^{n-1}$. –  John Klein Feb 24 '11 at 4:16

Alright, here is an idea for when $A$ is a skeleton; let's say $A=X^{n-1}\subseteq X$ is the $(n-1)$-skeleton. For now assume that $X$ is $n$-dimensional, the statement for arbitrary $X$ should follow by induction.

From my understanding, $f$ and $g$ are homotopic relative $A$ if and only if their difference cochain $d(f,g)\in C^n(X;\pi_n(Y))$ vanishes (cf. Mosher and Tangora, Theorem 4). So let's assume that $d(f,g)\neq 0$, that is $f$ and $g$ are not homotopic relative $A$, and proceed with all the assumptions made above.

Since $B$ is assumed to be a contractible subcomplex of $Y$, the projection $pr:Y\to Y/B$ is a homotopy equivalence and induces an isomorphism $\pi_n(pr):\pi_n(Y)\to\pi_n(Y/B)$. Furthermore, we also get an induced map $p_\ast:C^n(X;\pi_n(Y))\to C^n(X;\pi_n(Y/B)),\ \phi\mapsto\pi_n(pr)\circ\phi$, which is injective by the injectivity of $\pi_n(pr)$. Thus $p_\ast(d(f,g))\neq 0$, where by the functoriality of $\pi_n(-)$ we (should) have $p_\ast(d(f,g))=d(pr\circ f,pr\circ g)\in C^n(X;\pi_n(Y/B))$. However, $d(pr\circ f,pr\circ g)=0$ since $pr\circ f$ and $pr\circ g$ are assumed to be homotopic relative $A$, which contradicts our assumption of $f$ and $g$ not being homotopic relative $A$.

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