homomorphisms over direct sum I am trying to compute some homology groups and for that I need to figure out,
what are all homomorphisms from $\mathbb Z \oplus \mathbb Z$ into $\mathbb Z$. 
I would really appreciate any effort.
 A: $\mathbf Z \oplus \mathbf Z \cong \mathbf Z^2$ is a free abelian group. So you can choose the images of the standard base vectors $e_1=(1,0)$ and $e_2=(0,1)$ of $\mathbf Z^2$ under a homomorphism in any abelain group in an arbitrary way. So any homomorphism from $\mathbf Z^2$ to $\mathbf Z$ is a follows:
$$
\varphi( x_1 e_1+x_2 e_2)=x_1 k + x_2 m
$$
where $k,m $ are fixed integers, the images of $e_1$ and $e_2,$ respectively. 
A: Hint: It is enough to determine the image of the generators.  If you know
$$
\varphi(1,0)=a\qquad\text{and}\qquad\varphi(0,1)=b,
$$
then you can define
$$
\varphi(x,y)=\varphi((x,0)+(0,y))=\varphi(x,0)+\varphi(0,y)=x\varphi(1,0)+y\varphi(0,1)=xa+yb.
$$
This describes all homomorphisms and every such map is a homomorphism. 
If $a$ and $b$ are both zero, this is the zero map and the kernel is all of $\mathbb{Z}\oplus\mathbb{Z}$.  Suppose that $a$ and $b$ are not both zero. Let $g=\gcd(a,b)$ so that $a=a'g$ and $b=b'g$.  Then, the kernel of $\mathbb{Z}\oplus\mathbb{Z}=(b',a')\mathbb{Z}$.  Therefore, none of these maps are isomorphism (there are many other ways to argue that these groups are not isomorphic).
A: The general answer is that, if $R$ is a commutative ring,  $F$ is a free $ R$-module: $F\simeq R^{(I)}\;$ for some set $I$, then for  any $R$-module $M$, one has
$$\operatorname{Hom}_R(F,M)\simeq M^I.$$
In particular,
$$\operatorname{Hom}_\mathbf Z(\mathbf Z^2, \mathbf Z )\simeq \mathbf Z^2.$$
If the elements of $\mathbf Z^2$ are represented by column matrices of $\mathcal M_{2\times1}(\mathbf Z)$, the homomorphisms of  $\mathbf Z^2$ into  $\mathbf Z $ are represented by row matrices of  $\mathcal M_{1\times2}(\mathbf Z)$.
