Category Kernel Question (Unique homomorphism $\alpha_0:\ker\phi\to L$) There is a a question with two parts (prove or disprove), one of which I know how to do, but the other part I am stuck.
Let $\phi:M'\to M$ be a homomorphism of abelian groups. Suppose that $\alpha:L\to M'$ is a homomorphism of abelian groups such that $\phi\circ\alpha=0$, for instance the inclusion map $\mu:\ker\phi\to M$. Prove or disprove each of the following:

*

*There is a unique homomorphism $\alpha_0:\ker \phi\to L$ such that
$\mu=\alpha\circ\alpha_0$.
(This I am unsure if it is true or untrue)


*There is a unique homomorphism $\alpha_1:L\to\ker\phi$ such
that $\alpha=\mu\circ\alpha_1$.
(This I am quite sure it is true as it is the universal property of the kernel.)
I will post my solution to part 2 below.

Part (ii) is true.
Uniqueness:
Suppose $\alpha=\mu\circ\alpha_1=\mu\circ\beta$. Then $\mu\alpha_1(x)=\mu\beta(x)$.
This means that $\mu(\alpha_1(x)-\beta(x))=0$.
Hence $\alpha_1(x)-\beta(x)=0$ and $\alpha_1(x)=\beta(x)$.
Existence: Take $\alpha_1:L\to \ker\phi$, $\alpha_1(l)=\alpha(l)$.
Since $\phi(\alpha(l))=0$, thus $\alpha(l)\in\ker\phi$.
$\alpha_1(l_1+l_2)=\alpha(l_1+l_2)=\alpha(l_1)+\alpha(l_2)=\alpha_1(l_1)+\alpha_1(l_2)$ hence it is a homomorphism.
$\mu\alpha_1(l)=\mu\alpha(l)=\alpha(l)$.
