A map between spaces induces maps between homology groups and cohomology groups (in the opposite direction). How are these induced maps related?
I'm currently working through Hatcher. My intuition is that the universal coefficient theorem gives us a natural structure I'd want: I expect that the action that $f$ induces on $H_n(X) \to H_n(Y)$ is reflected in $Hom(H_n(X),G) \to Hom(H_n(Y),G)$, and that that is then reflected in the appropriate factors on either side of $H^n(Y) \to H^n(X)$. And that similarly the action that $f$ induces on $H_{n-1}(X) \to H_{n-1}(Y)$ is reflected in $Ext(H_{n-1}(X),G) \to Ext(H_{n-1}(Y),G)$, and that that is then reflected in the appropriate factors on either side of $H^n(Y) \to H^n(X)$.
However, I can't find a clear reference to this statement in Hatcher. Is it there and I'm missing it? I'm generally bad at noticing when the book makes these 'natural' statements explicit. Or is there a reference somewhere else? Or am I way off base?
You can find this in Hatcher after Corollary 3.3 on page 196 for abstract chain complexes and in the middle of page 201 for cohomology of spaces. In particular, the short exact sequence of the universal coefficient theorem to compute the cohomology of a space $X$ is natural in $X$. This means that if $f:X\to Y$ then the diagram $$\require{AMScd} \begin{CD} 0 @>>>\operatorname{Ext}(H_{n-1}(Y),G)@>{}>> H^n(Y;G) @>{}>> \operatorname{Hom}(H_{n}(Y),G) @>>>0\\ & @V{}VV @V{}VV @V{}VV \\ 0 @>>>\operatorname{Ext}(H_{n-1}(X),G)@>{}>> H^n(X;G) @>{}>> \operatorname{Hom}(H_{n}(X),G) @>>>0 \end{CD}$$ commutes where the vertical maps are induced by $f$. Explicitly, the middle vertical map is the induced map $f^*$ on cohomology. The right vertical map is the map given by composition with the induced map $f_*:H_n(X)\to H_n(Y)$. Finally, the left map is the map induced by $f_*:H_{n-1}(X)\to H_{n-1}(Y)$ using the fact that $\operatorname{Ext}(-,G)$ is a contravariant functor.