# 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.

• +1 for the "double" proof. Down to earth and explained in homological algebra terms. Commented Feb 29, 2016 at 10:01
• Hi, I know I left this as answered but I would like to clarify a few points. To prove it is an isomorphism, I must prove that $ker \phi = 0$ and $coker \phi = 0.$ $ker \phi = 0$ is fairly simple, but to show $coker \phi = 0,$ I need to show that $B/Im(\phi) = 0,$ but it seems to me that $Im(\phi) = Im(\theta_2 + s) = Im(\theta) + Im(s) = A \oplus b\mathbb{Z}$ which only equals B if b=1, and so is therefore not an isomorphism? Thank you for your help so far! Commented Mar 1, 2016 at 16:20