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I'm having some trouble figuring out exactly how to prove that an inclusion map induces an isomorphism on homology.
Let X be the 1-skeleton of the Torus $T$, so a wedge of two circles. Let $i:X\rightarrow T$ be the inclusion map, and let $i_*:H_1(X)\rightarrow H_1(T)$ be the induced inclusion. I wish to show this is an isomorphism.
I know the homology for both spaces, so I know $H_1(X)=H_1(T)=\mathbb{Z}\oplus \mathbb{Z}$. I also know that $i_*$ is injective, so all I need is surjectivity.
My attempt at a solution used relative homology to get a long exact sequence: $$H_2(X)\rightarrow H_2(T)\rightarrow H_2(T,X)\rightarrow H_1(X)\rightarrow H_1(T)\rightarrow H_1(T,X)$$ $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow\mathbb{Z}\oplus \mathbb{Z}\rightarrow 0$$ Where the last nonzero map is $i_*$; but I don't think that gives me anything I didn't have before. Any help is appreciated!

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    $\begingroup$ What does exactness at $H_1(T)$ tell you? $\endgroup$ – aes Mar 6 '15 at 4:22
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    $\begingroup$ You mean that $X$ is the 1-skeleton, not 2-skeleton, yes? $\endgroup$ – Neal Mar 6 '15 at 4:25
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Hint 1: As aes asks in the comments, what does exactness at $H_1(T)$ say?

Hint 2: The map $\mathbb{Z}\to\mathbb{Z}$ has trivial kernel. Because $\mathbb{Z}\oplus\mathbb{Z}$ has no torsion, the map $\mathbb{Z}\to\mathbb{Z}$ is an isomorphism. Now what does exactness say?

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    $\begingroup$ Okay, I think I was just having a brain fart. Exactness at $H_1(T)$ gives surjectivity right away, and since $\mathbb{Z}\rightarrow\mathbb{Z}$ is an isomorphism, it has an image of $\mathbb{Z}$, which means the map from $\mathbb{Z}\rightarrow\mathbb{Z}\oplus \mathbb{Z}$ is the zero map by exactness, so $i_*$ has a trivial kernel by exactness. $\endgroup$ – TinaBelcher Mar 6 '15 at 4:49
  • $\begingroup$ @TinaBelcher Yes, although it may be faster to observe that surjectivity between two free abelian groups of the same rank implies isomorphism. $\endgroup$ – Neal Mar 6 '15 at 12:41

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