Hom 2

Statement I is FALSE : Take $f:\mathbb{Z}\hookrightarrow\mathbb{Q}$ the natural inclusion, so that $G_1=\mathbb{Z}$ and $G_2=\mathbb{Q}$ . Furthermore take $H=\mathbb{Z}/2$ . Then you obtain the following sets:

$\mathrm{Hom}(G_2,H)=\{$ trivial map $\}$

$\mathrm{Hom}(G_1,H)=\{$ trivial map, natural projection, according to where $1\in\mathbb{Z}$ is sent in $H=\mathbb{Z}/2\}$

Thus the map $\tilde{f}:\mathrm{Hom}(G_2,H)\rightarrow\mathrm{Hom}(G_1,H)$ can't be surjective.

Statement II is TRUE : It suffices to show that $\mathrm{Ker}(\tilde{f})=\{0\}$ since it's a group homomorphism. So take $h:G_2\rightarrow H$ such that $h\circ f =0$ the zero map. Let $x\in G_2$ ; since $f$ is surjective we can find a $y\in G_1$ s.t. $f(y)=x$ so:

$h(x)=h\circ f(y)=0$ Since the $x$ was arbitrary, $h=0$ .

The converse of I is TRUE : Simply take $H=G_1$ . Since $\tilde{f}$ is surjective, there is $g:G_2\rightarrow G_1$ s.t. $g\circ f=Id_{G_1}$ thus $f$ is injective.

The converse of II is TRUE : Set $C=\mathrm{Coker}(f)$ and $\pi :G_2\rightarrow C$ the natural projection, so that $\pi\circ f = 0$ , i.e. $\tilde{f}(\pi)=0$ . Since $\tilde{f}$ is supposed to be injective, one has $\pi =0$ which amounts to $C=0$ since $\pi$ is surjective.

Remark, for the people who know some category theory : all of this could be formulated in terms of abelian categories. $\mathcal{Ab}$ , the category of abelian groups is abelian (or more generally $A-\mathcal{Mod}$ the category of $A$ -modules, for a ring $A$ ), so the fact that morphisms are epi or mono can be reformulated in terms of kernels and cokernels. Showing that the functors $\mathrm{Hom}(-,H)$ are left-exact $\forall H$ abelian groups (or more generally, any $A$ -module) uses arguments such as the ones given to prove statement II and its converse. We just need to take the image of the following exact sequence: $G_1\rightarrow G_2\rightarrow \mathrm{Coker}(f)\rightarrow 0$ through $\mathrm{Hom}(-,\mathrm{Coker}(f))$ . We sometimes write $\mathrm{Hom}(f,H)$ for the morphism $\tilde{f}$ .

This is simply reformulating the problem with a bit of height and doesn't add much to the way we solve the exercise, but it helps to figure out where we are headed. We could have done all of this in a general abelian category.