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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.

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    A linear map is determined by its action on a basis. Can you use a similar idea here? – 2017-02-12

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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).

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    No actually it is isomorphic to $\mathbb Z$ if $a=b$ or $a=-b$, am I right? But what about other cases, since I don't know what a and b are (I only have general notion of homomorphism in my exercise) – 2017-02-12
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    $\mathbb{Z}\oplus\mathbb{Z}$ is not isomorphic to $\mathbb{Z}$. None of these maps are isomorphisms (nor can they be). The kernel is the set of all $x,y$ such that $xa+yb=0$. – 2017-02-12
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$\mathbf Z \oplus \mathbf Z \cong \mathbf Z^2$ 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.

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    Thanks. But does this mean that the kernel of any such homomorphism is trivial? – 2017-02-12
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    No! Were the kernel trivial, $\mathbf Z^2$ would be isomorphic to a subgroup of $\mathbf Z$ (why?) It's certainly not the case. Say, in my notation above, any element of $\mathbf Z^2$ of the form $l(m e_1-k e_2)$ goes to zero under $\varphi.$ – 2017-02-12
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    I apologize, my mistake. Actually it is isomorphic to $\mathbb Z$ if $a=b$ or $a=−b$, am I right? But what about other cases, since I don't know what a and b are (I only have general notion of homomorphism in my exercise) – 2017-02-12
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    Aham, under this notation and if an element of the form $l(me_1-ke_2)$ goes to zero, does this mean that the kernel is isomorphic to $\mathbb Z$ in any case? – 2017-02-12
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    The kernel is isomorphic to $\mathbf Z,$ but the image isn't. Rougly speaking, two groups are isomorphic iff they have the same algebraic properties. $\mathbf Z$ is a cyclic group, while $\mathbf Z^2$ isn't. In fact, ANY homorphism from $\mathbf Z^2$ into $\mathbf Z$ has a nontrivial kernel. – 2017-02-12
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    But what if both $a$ and $b$ are zero? Then obviously kernel is not isomorphic to $\mathbb Z$ – 2017-02-12
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    In this case, all elements of $\mathbf Z^2$ go to zero, you have the trival homomorphism, the kernel is equal to $\mathbf Z^2.$ In your terms: whenever one of $a,b$ is nonzero, the kernel is a cyclic group of $\mathbf Z^2.$ Otherwise, the kernel is equal to the whole of the group. Right? – 2017-02-12
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    Yes. Just one question left.. What are all cyclic groups of $\mathbb Z^2$? – 2017-02-12
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    Any group: take an element, generate all its powers (multiples). The group $\mathbf Z^2$ is no differenent: $$ \langle x_1 e_1 + x_2 e_2 \rangle =\{ k x_1 e_1 + k x_2 e_2 : k \in \mathbf Z\}. $$ So you'll have either infinitely many elements, or just one if $x_1=x_2=0.$ – 2017-02-12
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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 $\mathcal M_{1\times2}(\mathbf Z)$.