# Relative Simplicial Approximation

While I was studying the cellular approximation theorem on May's "A Concise Course in Algebraic Topology" I found something a bit unclear. I agree with the fact that, given two CW-complexes $X,Y$, and a map $f:X \to Y$, $f$ is homotopic to a simplicial map, essentially because there is an equivalence of category $Ho(SSet) \simeq Ho(K)$, given by the geometric realization functor, where $K$ is the category of compactly generated weak Hausdorff spaces.

But the author uses a "relative" version, when considering a relative CW-complex $(X,A)$ (i.e. a space obtained by $A$ by adding cells, where $A$ is arbitrary): we consider a map $f:(I^q, \partial I^q, J^q) \to (X,A,a)$ representing an element of $\pi_q (X,A,a)$, as usual. He then claims that this map is homotopic $rel\ \partial I^q$to a simplicial map $f'$ which agrees with $f$ on $\partial I^q$.

Why is that true? Going through Spanier and n-lab I've seen that this holds when $f$ is already simplicial on $\partial I^q$, but in principle $A$ is not endowed with a simplicial structure!

Any hint is welcome, thanks in advance!

Actually, I'm able to solve this problem when $A$ is a subcomplex of $X$, by first approximating $f_{|\partial I^q}$ to be simplicial, then using the fact that $\partial I^q \to I^q$ is a cofibration I can extend this homotopy obtaining $\tilde{f}:I^q \to X$, simplicial on $A$, hence I can conclude by the relative simplicial approximation.
• Take also a look at Theorem 4.8 at page 349 in Hatcher's algebraic topology. If your map is already cellular on your subcomplex $A$, you can further require the homotopy to be stationary on $A$. – Mauro Porta May 13 '14 at 12:41