The setup of the question is as follows. Let the following be a short exact sequence of modules:
$$ 0 \rightarrow A \xrightarrow{i} B \xrightarrow{p} C \rightarrow 0, $$ and let $r: B\to A$ be a map such that $r \circ i = id_A$. We want to show that this implies the sequence splits, i.e., that $B\cong A \oplus C$.
This is essentially one part of the splitting lemma (for modules). The site nCatLab gives a proof of this (for abelian categories more generally), which goes as follows:
Conversely, suppose we have a retract $r: B \rightarrow A$ of $i: A \rightarrow B$. Write $P: B \rightarrow A \rightarrow B$ for the corresponding idempotent. Then every element $b \in B$ can be decomposed as $b=(b-P(b))+P(b)$ hence with $b-P(b) \in \operatorname{ker}(r)$ and $P(b) \in \operatorname{im}(i)$. Moreover this decomposition is unique since if $b=i(a)$ while at the same time $r(b)=0$ then $0=r(i(a))=a$. This shows that $B \simeq \operatorname{im}(i) \oplus \operatorname{ker}(r)$ is a direct sum and that $i: A \rightarrow B$ is the canonical inclusion of $\operatorname{im}(i)$. By exactness it then follows that $\operatorname{ker}(r) \simeq \operatorname{im}(p)$ and hence that $B \simeq A \oplus C$ with the canonical inclusion and projection.
Question. I have a question about one step of this proof. They write "By exactness it then follows that $\operatorname{ker}(r) \simeq \operatorname{im}(p).$" My question is: how exactly (no pun intended!) does this statement follow from exactness?
My understanding of exactness of the sequence is just that $\operatorname{im}(i) = \operatorname{ker}(p)$ (along with $i$ being injective and $p$ being surjective). But how would exactness imply anything about $\operatorname{ker}(r)$?