Let $\mu$ be a homomorphism and $\alpha$ an endomorphism with appropriate domains and codomains.

Suppose $\alpha (\ker \mu) \subseteq \ker \mu$. Why does this imply that there is some endomorphism $\beta$ such that $\beta \circ \mu = \mu \circ \alpha$?

This seems like something involving the universal property of the kernel.

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    $\begingroup$ What's the context? Groups, modules, algebras? $\endgroup$ – egreg Aug 4 '16 at 20:56

By the given inclusion, $\mu\circ\alpha$ is $0$ on $\ker\mu$, i.e., $\ker\mu\subseteq \ker(\mu\circ\alpha)$. Hence $\mu\circ\alpha$ factors over $\mu$.


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