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.



About mathtuition88

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