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I should prove that:

if I have the short exact sequence ${0}\rightarrow A \rightarrow B \rightarrow C \rightarrow {0}$ with $f$ injective homomorphism between $A$ and $B$ and $g$ surjective homomorphism between $B$ and $C$, and $C$ is an abelian free group, then $B$ is isomorphic to the direct product of $A$ and $C$.

I've been suggested to use (at least if $C=Z$, don't know if it works in general) the homomorphism $h(a,c)= f(a)+ \phi(c)$, where $\phi(1)=b$ and $b \in B$ such that $g(b)=1$. It is said that this is an isomorphism, but I don't see how to prove that it is injective and surjective.

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    A hint for injectivity: show that it implies $f(x) = \phi(y)$ for some $x,y$, and apply $g$ to both sides. What do you know about $gf$? About $g\phi$? – 2017-02-06
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    Exactness says a lot about $gf$, and by construction you can compute $g\phi(x)$ for any $x$ using the defining property of homomorphism. In particular, when does $g\phi(x)$ vanish? – 2017-02-06

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Let $\{c_i\}$ denote a set of generators of the free abelian $C$. To define $\phi:C \to B$, it suffices to define $\phi$ over the elements $c_i$, then extend this to a homomorphism over all of $C$ by "linearity". In particular, we define $\phi(c_i)$ is such that $g(\phi(c_i)) = c_i$. This property extends to all of $C$, which is to say that $g(\phi(c)) = c$ for any $c \in C$. We then define $$ h(a,c_i) = f(a) + \phi(c_i) $$

Note: we can guarantee that the conditions on $\phi$ define a homomorphism since $C$ is a free abelian group.

For injectivity: suppose that $h(a,c) = 0$. Then $$ f(a) + \phi(c) = 0 \implies f(a) = -\phi(c) $$ However, $f(a)$ is in the kernel of $g$. Thus, applying $g$ to both sides produces $0 = -g(\phi(c)) = c$. That is, $c = 0$. Thus, $h(a,c) = f(a) + \phi(0) = f(a)$, and so $f(a) = 0$. But, $f$ is injective, so $a = 0$.

Surjectivity: note that $g$ induces an isomorphism $\tilde g:B/\ker g \to C$. Thus, for every $b \in B$, there exists a $c \in C$ such that $$ \tilde g(b + \ker(g)) = c \implies \tilde g(b+\ker g) = \tilde g(\phi(c) + \ker g) \implies\\ b - \phi(c) \in \ker(g) $$ However, $\ker(g) = im(f)$. So, there exists an $a$ such that $b - \phi(c) = f(a)$. That is, $$ b = f(a) + \phi(c) = h(a,c) $$

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Let $ a_1,a_2,..a_n$ be the generators of C.Let $ b_1,b_2,...b_n$ be some preimages. Define $ j: C \to B$ by $ j(a_k) = b_k$. Now $ j \circ g = 1_C$. This is called split surjective. Now for this you shall get $ B \cong \ker g \oplus C \cong im f \oplus C \cong A \oplus C$

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    "This is called split surjective". What are you calling split surjective, exactly? "Now for this you shall get [isomorphisms]": of course he shall, but that doesn't do much to help with the proof. – 2017-02-06
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    This is a great answer if you're familiar with proving the splitting lemma, but in that case you probably wouldn't be asking this question. An important detail would be to include that $B$ is the direct sum of the kernel of $g$ and the image of $j$, and why that is the case. – 2017-02-06