I need to show that if we let $T_n(X,A)$ denote the torsion subgroup of $H_n(X,A)$, then the functor $(X,A)\mapsto T_n(X,A)$ with the obvious induced homomorphisms and boundary maps do not define a homology theory.
Using the axioms that Hatcher gives us, my guess is to find maps
$\tilde{H}_n(X/A)\rightarrow \tilde{H}_{n-1}(A)$
$\downarrow\hspace{3 cm}\downarrow$
$\tilde{H}_n(Y/B)\rightarrow \tilde{H}_{n-1}(B)$.
such that either $\tilde{H}_{n-1}(A)$ has zero torsion or $\tilde{H}_n(Y/B)$ has zero torsion, but not both. Then, the diagram for $T_i$ will not be commutative (I also have to make sure there aren't any trivial maps). The simplest map I can think of would be
$\mathbb{Z}_2\rightarrow \mathbb{Z}$
$\downarrow\hspace{1 cm}\downarrow$
$\mathbb{Z}_2\rightarrow \mathbb{Z}_2$.
but I don't know any spaces that have $H_i=\mathbb{Z}_2$ for $i$ even (at least Hatcher didn't go over any such spaces before this question). I just want to know if this is the right path to take, or if I'm wrong and commutativity does, in fact, hold, and I should consider something else.