I was wondering if the following were true or not. Suppose $f:(X,X^n)\rightarrow (X,X^n)$ and $g:(X,X^n)\rightarrow (X,X^n)$ are homotopic. Does it neccesarily follow that $f$ and $g$ are homotopic rel $X^n$.
The reason I am asking is that I am trying to prove that the $n$-skeletons of homotopy equivalent CW complexes without cells of dimension $n+1$ are also homotopy equivalent. What I have so far is the following. Note that this is exercise 14 page 359 in Hatcher.
Proof: Let $f:X\rightarrow Y$ be a homotopy equivalence and let $g:Y\rightarrow X$ be its counterpart. Then we can homotope these maps to get cellular maps $\tilde{f}$ and $\tilde{g}$ ,respectively. In particular $\tilde{f}:(X,X^{n})\rightarrow(Y,Y^{n})$ and $\tilde{g}:(Y,Y^{n})\rightarrow(X,X^{n})$ . And furthermore, we have $\tilde{f}\circ\tilde{g}\approx f\circ g\approx\text{Id}_{Y}$ and $\tilde{g}\circ\tilde{f}\approx g\circ f\approx\text{Id}_{X}$ . The key thing we need is to ensure that the homotopies are taken relative $X^{n}$ and $Y^{n}$ . Then taking restrictions on our maps gives us the desired result. I'm assuming that I have to work in the fact that the $CW$ complexes don't contain $n+1$ cells, but I don't see how that gets worked into it.
No, this isn't true in general. For instance, if $X=X^n$, then you are just asking whether any two homotopic maps from a $\leq n$-dimensional CW-complex to itself are equal, which is obviously false for most such $X$.