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



Lifting relative homotopies between maps $X\to Y/B$ when $B$ is contractible
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 $\pi\_0$.


