0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ What's the context? Groups, modules, algebras? $\endgroup$
    – egreg
    Aug 4, 2016 at 20:56

1 Answer 1

1
$\begingroup$

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 $\mu$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.