For context I am working on Weibel's K-book, trying to understand theorem 4.4.3. For this I am reading the paper by Caruso, Cohen, May and Taylor that Weibel refers to. There is a key lemma used in the proof which is:
Let $f:X\to X'$ be an integral homology isomorphism, and $Y$ a group like space. Then $f$ induces an isomoprhism $[X', Y]\to [X, Y]$ and also an isomorphism $[X',Y]_w\to [X,Y]_w$, where this notation means weak homotopy classes of maps. To recall we call two maps weakly homotopic if the precomposition by any map from a finite CW complex yields homotopic maps.
I have succeeded in understanding the proof of the case $[X',Y]\to [X,Y]$, and I think surjectivity in the weakly homotopic case can easily be deduced from this. Hence I have injectivity left to show. We can rephrase what we want to show as follows, suppose $g:X'\to Y$ is such that $g\circ f: X\to Y$ is weakly homotopic to a trivial map, then $g$ is weakly homotopic to a trivial map.
So let $i': A\to X'$ be a map from a finite CW complex, we want to show $g\circ i'$ is (not weakly) nullhomotopic. If we can lift the map $i$ to a map $i: A\to X$ such that $f\circ i=i'$ we are done, because $g\circ f$ is weakly nullhomotopic.
Had $f$ been a weak equivalence (homotopy group isomorphism) this wouldn't be hard, I naively hope that $f$ being an integral homology isomorphism is enough to lift $i'$, operating one cell at a time because there are only finitely many cells to lift. Would anybody know how to this?
If this isn't possible, the natural guess is to somehow use that $Y$ is a group like space, but I haven't had any ideas on how to use this assumption.
Perhaps if I was more comfortable with obstruction theory I would be more comfortable with this, but my knowledge of that subject is sadly limited.
Thanks a lot to anyone willing to help.
I think I have an answer.
We can make $f$ cellular and then a cofibration, in order to assume that $f$ is a cellular inclusion $X\subset X'$. By the homology long exact sequence we have $H_*(X', X; \pi)=0$. Let $\phi: X'\to Y$ which when restricted to $X'$ is weakly null homotopic. We want to show $\phi$ is weakly null homotopic. Let $A'\subset X'$ be a finite CW complex. In order to apply excision, we extend $A'$ to $A''$ by adding every cell in $X'$ which isn't in $X$. Now Let $A=A''\cap X=A'\cap X$ which is a finite sub complex of $X$. So $\phi|_A\simeq *$ by assumption. Now the obstruction to extending this homotopy to $A''$ lies in $H_n(A'', A; \pi_n(Y))$ because every space involved is a CW complex and $Y$ is a simple space. But by excision (which can be applied as $X'=X\cup A''$ is a union of subcomplexes) we have that $H_n(A'', A;\pi_n(Y))=H_n(X', X;\pi_n(Y))=0$, so the homotopy can be extended to $A''$, in particular to $A'$. This proves that $\phi$ is weakly homotopic to a trivial map as desired