Splitting Lemma where $C=\mathbb{Z}.$ Given a short exact sequence
$$ 0 \xrightarrow{\theta_3} A \xrightarrow{\theta_2} B \xrightarrow{\theta_1} \mathbb{Z} \xrightarrow{\theta_0} 0 $$
show that $B \cong A \oplus \mathbb{Z}.$
So far I have that $\theta_2$ is injective and as $0 \to \operatorname{Im}(\theta_3) \to A \to \operatorname{Im}(\theta_2) \to 0 $ is exact, then $A \cong Im(\theta_2).$ Similarly, $\operatorname{Im}(\theta_1) \cong \mathbb{Z}$ from the surjectivity of $\theta_2.$
Also from exactness we have that $\operatorname{Im}(\theta_2) \cong \operatorname{Ker}(\theta_1).$

This is the elementary way to proof it:
Let $b \in B$ with $\theta_1(b)=1$ and define a homomorphism $s: \mathbb Z \to B, 1 \mapsto b$.
Then you can show that 
$$A \oplus \mathbb Z \to B, (a,z) \mapsto \theta_2(a) + s(z)$$
is an isomorphism.

Using results from homological algebra, one would just say that $\mathbb Z$ is free, hence projective. Thus the sequence splits and we obtain the result.
