abelian groups – Mathtuition88

Recently, the Dayan Zhanchi brand Rubik’s Cube has been a hot sale on my site. Parents looking to buy an educational toy may want to check it out. Smoothness when turning is very important for Rubik’s Cube, and the Dayan Zhanchi is one of the smoothest cubes out there (much better than the ‘official’ brand).

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 $\alpha(l)\in\ker\phi$ .