## Universal Property of Kernel Question

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In this blog post, we will discuss a category theory question, in the framework of homomorphisms of abelian groups.

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 $\phi\circ\alpha$ is the zero map. (One example is the inclusion $\mu:\ker\phi\to M'$)

Are the following true or false?

(i) There is a unique homomorphism $\alpha_0:\ker\phi\to L$ such that $\mu=\alpha\circ\alpha_0$.

(ii) There is a unique homomorphism $\alpha_1:L\to\ker\phi$ such that $\alpha=\mu\circ\alpha_1$.

It turns out that (i) is false. We may construct a trivial counterexample as follows. Consider $L=M=0$, and $M'=\mathbb{Z}/2\mathbb{Z}$. Let $\alpha$, $\phi$ be both the zero maps. Then certainly $\phi\circ\alpha=0$. $\ker\phi=\mathbb{Z}/2\mathbb{Z}$. Then, for any $\alpha_0$, $\alpha\circ\alpha_0(x)=0$, and hence is not equals to the the inclusion map $\mu$.

It turns out that (ii) is true, in fact it is the famous universal property of the kernel, that any homomorphism yielding zero when composed with $\phi$ has to factor through $\ker\phi$.

First we will prove uniqueness. Let $\alpha=\mu\circ\alpha_1=\mu\circ\beta$, where $\beta$ is another such map with the property (ii). Then for all $x\in L$, $\mu\alpha_1(x)=\mu\beta(x)$, which implies $\mu(\alpha_1(x)-\beta(x))=0$. Since $\mu$ is the inclusion map, this means that $\alpha_1(x)-\beta(x)=0$ and thus $\alpha_1(x)=\beta (x)$.

Next, we will prove existence. Consider $\alpha_1:L\to\ker\phi, \alpha_1(l)=\alpha(l)$. Note that $\phi(\alpha(l))=0$ by definition thus $\alpha(l)\in\ker\phi$.

Next we prove it is a homomorphism. $\alpha_1(l_1l_2)=\alpha(l_1l_2)=\alpha(l_1)\alpha(l_2)=\alpha_1(l_1)\alpha_1(l_2)$.

Finally by construction it is easy to see that $\mu\alpha_1(l)=\mu\alpha(l)=\alpha(l)$ for all $l\in L$.