Lee's proof of the rank theorem for abelian groups I am going through Prof. Lee's "Introduction to Topological Manifolds" second time through, trying to do all the exercises and problems. My question is about a proof of the rank theorem for finitely generated abelian groups. If it is trivial or silly, I will gladly delete it. I just want to go through the book carefully, dotting the i's and crossing the t's.
Suppose $f:G\to H$ is a surjective homomorphism of abelian groups. Assume that $G$ is finitely generated and let $K=\ker f$.  Then $\text{rank}\ G= \text{rank}\ K+ \text{rank}\ H.$
Recall that the rank of an abelian group $G$ is defined by $\text{rank}\ G:=\text{rank}\ G/G_{tor}.$
One has the homomorphism $\tilde f:G/G_{tor}\to H/H_{tor}$ defined in the obvious way.
Now, Lee states "clearly the kernel of $\tilde f$ contains $K/K\cap G_{tor}".$ I don't understand what is meant by this. How are $\ker \tilde f$ and elements of $K/K\cap G_{tor}$ comparable? Or are we considering $\tilde f$ "restricted" to the latter set?
I think I can do the proof in a different way. A sketch:
If we consider the short exact sequence $0\rightarrow \ker \tilde f\xrightarrow{\tilde i} G/G_{tor}\xrightarrow{\tilde f} H/H_{tor}\rightarrow 0$,  since each of these groups is free abelian, the sequence can be written $0\to \mathbb{Z}^{n}\to \mathbb{Z}^{m}\to \mathbb{Z}^{l}\to 0$ from which it follows that $\mathbb{Z}^{m}\simeq \mathbb{Z}^{n}\oplus \mathbb{Z}^{l}$ because the sequence splits. The result follows from this.
But of course, this is not the way Lee does the proof.
edit: $\ker \tilde f$ is certainly free abelian, but as pointed out in the answer, relating it to $K$ itself is just Lee's approach.
 A: How is $\tilde{f}$ defined? You have the map $f\colon G\to H$, and you have the map $\pi\colon H\to H/H_{\mathrm{tor}}$. Since $f(G_{\mathrm{tor}})\subseteq H_{\mathrm{tor}}$, the kernel of $\pi\circ f$ contains $G_{\mathrm{tor}}$, so it factors through $G/G_{\mathrm{tor}}$; this is the map $\tilde{f}$.
What is the kernel of $\pi\circ f$? It contains $K$, and it contains $G_{\mathrm{tor}}$. Therefore, it contains $KG_{\mathrm{tor}}$. Thus, the kernel of $\tilde{f}$ contains $KG_{\mathrm{tor}}/G_{\mathrm{tor}}$.
By the Second (or Third, depending how you number them) Isomorphism Theorem,
$$\frac{KG_{\mathrm{tor}}}{G_{\mathrm{tor}}} \cong \frac{K}{K\cap G_{\mathrm{tor}}}.$$
So Lee is abusing language a bit by saying the kernel "contains" $K/K\cap G_{\mathrm{tor}}$. It actually contains the canonically isomorphic group $KG_{\mathrm{tor}}/G_{\mathrm{tor}}$.

Your proposed path does not work. You say that the short exact sequence
$$1\to K\to G\to H\to 1$$
induces a short exact sequence
$$1 \to \frac{K}{K_{\mathrm{tor}}} \to \frac{G}{G_{\mathrm{tor}}} \to \frac{H}{H_{\mathrm{tor}}} \to 1.$$
But that sequence need not be exact: consider the case of $K=2\mathbb{Z}$, $G=\mathbb{Z}$, and $H=\mathbb{Z}/2\mathbb{Z}$. The sequence you get by moding out by the torsion is
$$1 \to 2\mathbb{Z} \to \mathbb{Z}\to 1\to 1,$$
with the first map the canonical inclusion. But this map is not surjective, which is what you would need for this sequence to be exact.
The problem is that while $G_{\mathrm{tor}}$ certainly maps to $H_{\mathrm{tor}}$, it need not be the only thing that maps to $H_{\mathrm{tor}}$: you can have nontorsion elements of $G$ become torsion in the quotient.
