Question: I've been trying to figure out this homework problem for a while now and always get stuck. Both groups have 2 elements of order 4, and the remaining non identity elements have order two. Since the image of the homomorphism is a subgroup of $Z\_4$ and any subgroup of $Z\_4$ has order 1,2, or 4, I'm trying to show that the image of any homomorphism between the two does not have order 4. Does that sound right? I'm not sure where to go after this. I think the key of this is that $D\_4$ has 2 generators, but only one is order 4.

Answer: 
Your idea is a good one, since the groups in question are small. Start with your generating reflection $r$ and rotation $\rho,$ and assume that $\varphi:D\_4\to\Bbb Z\_4$ is a homomorphism. Since the order of $\varphi(r)$ must divide the order of $r,$ then $\varphi(r)$ must have order $1$ or $2,$ and thus, we have either $\varphi(r)=0$ or $\varphi(r)=2.$ Neither $0$ nor $2$ generates $\Bbb Z\_4,$ so the only way to have surjectivity is if $\rho$ is sent to $1$ or $3.$

Seems like we're home free! But wait a moment...is it possible to have a homomorphism that does this? The answer, it turns out, is "no." Note that we would expect $$\varphi(\rho r)=\varphi(\rho)+\varphi(r)=\varphi(r)+\varphi(\rho)=\varphi(r\rho),$$ since $\varphi$ is a homomorphism and $\Bbb Z\_4$ is abelian. But $r\rho=\rho^3r,$ so we would need $$\varphi(\rho r)=\varphi(\rho^3r)=\varphi(\rho^2\rho r)=\varphi(\rho^2)+\varphi(\rho r).$$ But this can only happen if $$0=\varphi(\rho^2)=2\varphi(\rho),$$ which is only possible if $\varphi(\rho)$ is $0$ or $2.$
Assume there is such a homomorphism $\psi$. Then, $\psi$ turns some reflection $r$ to $0\in\mathbb{Z}\_4$ (otherwise, we would have $\psi(\rho)=0$ for every rotation $\rho$). It remains to note that there is another reflection $s$ (which has order two) such that $r,s$ generate $D\_4$.