[Math] Inclusion induces Isomorphism on Homology

Question

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!

Answer 1

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?