Homomorphism from $\mathbb{Z}\oplus \mathbb{Z}_2$ to $\mathbb{Z}$. What sort of homomorphisms can I have from $\mathbb{Z}\oplus \mathbb{Z}_2$ to $\mathbb{Z}$? What about if I know the homomorphism sends the $\mathbb{Z}$ part to zero. In other words, if I know my homomorphism sends (1,0) to 0, must my homomorphism be the zero homomorphism?
 A: Recall: According to the universal property defining the direct sum, a homomorphism of $A\oplus B$ to $C$ is "the same" as a homomorphism $A\to C$ and a homomorphism $B\to C$ combined. Here you picked a specific $A\to C$ and there is little choice for $B\to C$.
A: To your second question, yes-because $(0,1)$ must go to zero since its multiple by 2 does. Generally the maps are the same as those from $\mathbb Z$ to itself: $(1,0)$ goes to an rbitrary $n$ and $(0,1)$ goes to zero.
A: We obeserve that if we have a group $G$ and $N$ that are $p$,$q\in \mathbb{N}$, torsion (I.e. $g^p=1$ and $f^q=1$ for $g\in G$ and $f\in N$), for $(q, p)=1$, then find $m, k$ such that $mq+pk=1$. Now given a homomorphism $\phi:G\to N$, we have $\phi(g)=\phi(g^{mq+pk})=\phi(g)^{mq}\phi(g^{pk})=1^m\phi(1)^k=1$ so that the map is zero. Now if all elements of $N$ have infinite order (such as in $\mathbb{Z}$), we have $1=\phi(1)=\phi(g^q)=\phi(g)^q$ which implies $\phi(g)=1$ since all elements are of infinite order.
Now there is another important property about direct sums: it is called the universal property. What it say, is that if you have a direct sum $H=\bigoplus_i H_i$, $H_i$ abelian, then we have maps $\lambda_i:H_i\to H$, then given any homomorphism $\phi:H\to G$, $\bigoplus \phi\circ \lambda_i=\phi$. This says that we can decompose any mapping from a direct sum into the sum of mappings. (This becomes esspecially important when you start learning category theory, where we start calling them "coproducts", http://en.wikipedia.org/wiki/Coproduct )
Lets now solve your problem for a huge class of abeliain groups. If we have a group of the form $\mathbb{Z}^{r}\oplus \bigoplus_{p,\mathrm{prime}}\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}^{e_{p,n}}$ for some integers $r$, and $e_{p,n}$ mapping to $\mathbb{Z}^{q}\oplus \bigoplus_{p,\mathrm{prime}}\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}^{f_{p,n}}$. Then as above it, it suffices to consider the restrictions of the domain to each summand. We have that $\mathbb{Z}_{p^n}^{e_{p,n}}$ is $p^n$-torsion, so it cannot map nontrivially into any of the $\mathbb{Z}_{q^n}^{f_{q,n}}$ by the above aurguments. Thus it maps trivally into $\bigoplus_{q\neq q,\mathrm{prime}}\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{q^n}^{f_{q,n}}$. Also by above aurgument, it cannot map into $\mathbb{Z}$ nontrivally, so we have, summing together by the universal property, $\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}^{e_{p,n}}$ can only map into $\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}^{f_{p,n}}$ nontrivally. Sadly, we can say nothing about the free map, as seen by the projections $\mathbb{Z}\to \mathbb{Z}_n$. Thus we reach the final conclusion. That is given maps $\phi_i:\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}^{e_{p,n}}$ to $\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}^{f_{p,n}}$ (which are not to hard to classify) and a map $\psi:\mathbb{Z}^r\to \mathbb{Z}^{q}\oplus \bigoplus_{p,\mathrm{prime}}\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}^{f_{p,n}}$, we get a map $\psi\oplus\bigoplus \phi_i$ and all maps are of this form.
To have an example of what this means, take our initial example. $\mathbb{Z}\to \mathbb{Z}$ are easy, they are multiplication, and since our $\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{2^n}^{f_{p,2}}=0$ and $\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{2^n}^{e_{p,n}}=\mathbb{Z}_2$, we have no maps, and since we don't have to worry about any other components, the only maps we have ignore the $\mathbb{Z}_2$ part and are given by multiplication. I would advise any readers to work out generalizations and examples of these, there are quite a good exercise.
A: Consider that what you have done is essentially to specify the $\mathbb Z$ part as a subset of the kernel for your homomorphism. Now the $\mathbb Z$ part itself could be the kernel, in which case the only other coset left would be the $\mathbb Z$ part + (a, 0), which would get sent to some fixed (nonzero) element a of $\mathbb Z$. Now consider: could this be a homomorphism into $\mathbb Z$?
Of course, the only other option is for the kernel to be the whole group, and then the homomorphism is the zero map.
