TITLE: Inclusion induces identity on homology
QUESTION [0 upvotes]: Let $(H_*, \partial_*)$ be a homology theory with values in the category of $\Bbb{Z}$-modules satisfying the dimension axiom. Then the inclusion $S^1\vee S^1\to T^2$ should induce (up to isomorphism) the identity on $\Bbb{Z}\oplus \Bbb{Z}\cong H_1(S^1\vee S^1) \to H_1(T^2)\cong \Bbb{Z}\oplus \Bbb{Z}$ but I haven't been able to prove it.
I've tried using LES of the pair $(T^2, S^1\vee S^1)$ with no success. Note that this is in the setting of a general homology theory, and we may not use geometric considerations from singular homology.

REPLY [1 votes]: Assuming you're working with homology with coefficients in $\Bbb Z$, the long exact sequence of $(X, A) = (T^2, S^1 \vee S^1)$ works just fine : 
$$\cdots \to H_2(A) \to H_1(X) \to H_2(X, A) \stackrel{\partial}{\to} H_1(A) \to H_1(X) \to H_1(X, A) \to \cdots$$
As $H_2(A) = 0$, and $H_1(X, A) \cong H_1(X/A, \text{pt}) = H_1(S^2, \text{pt}) = 0$, this reduces to the exact sequence
$$0 \to \Bbb Z \to \Bbb Z \stackrel{\partial}{\to} \Bbb Z^2 \to \Bbb Z^2 \to 0$$
The second homomorphism has to be multiplication by some integer $n$. In that case, image has to be $n\Bbb Z$, which is in turn equal to kernel of $\partial$ by exactness. Thus, by first isomorphism theorem, $\Bbb Z/n\Bbb Z$ has to be a subgroup of $\Bbb Z^2$, which is possible if and only if $n$ is either $0$ or $1$, and the former is impossible as then the second map in the sequence would be zero, forcing $\Bbb Z \cong 0$.
Hence, the second map is an isomorphism, implying $\partial$ is the zero map. Thus, the fourth map $\Bbb Z^2 \to \Bbb Z^2$ has kernel $0$, and is thus injective. It's also surjective as the fifth map is zero, hence is an isomorphism.
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