TITLE: Kernel of morphism out of direct sum
QUESTION [3 upvotes]: Let $\mathcal{C}$ be an abelian category with objects $A,B,C \in \mathcal{C}$ for which we have a surjection $f:A\rightarrow C$ and an arbitrary morphism $g:B\rightarrow C$. Consider
\begin{align*} 
h:A\oplus B \stackrel{f+g}{\longrightarrow} C
\end{align*}
with $\ker(h)= \{(a,b) \in A \oplus B \mid f(a)=-g(b)\}$.
Intuitively I thought $\ker(h)=\ker(f)\oplus B$.
As $f$ ist surjective $f^{-1}(g(b))\neq \emptyset$ for every $b\in B$. Hence we have a surjection
\begin{align} 
\ker(h) \stackrel{\phi}{\rightarrow} B \\
(a,b) \mapsto b 
\end{align}
with $\ker(\phi)=\ker(f) \oplus (0)\cong \ker(f)$. Therefore we have a short exact sequence
\begin{equation*}
0 \rightarrow \ker(f) \rightarrow \ker(h) \rightarrow B \rightarrow 0. 
\end{equation*}
But so far I couldn't find any general splitting for this sequence. Is my intuition wrong?

REPLY [3 votes]: $A= \mathbb{Z}, B = \mathbb{Z}/2\mathbb{Z}. C = \mathbb{Z}/4\mathbb{Z}, f(n) = n\,(\mathrm{mod}\,4), g(m) = 2m$ gives a counterexample: the kernel of $h = f \oplus g$ is $\{(2k, k\,\mathrm{mod}\,2)\,|\, k\in \mathbb{Z}\}\simeq \mathbb{Z}$.
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