Prove that $f$ is an onto function and a homomorphism function from $(\mathbb{Z} \times \mathbb{Z}, \oplus)$ to $(\mathbb{Z}, +)$ I have a lot of issues trying to figure out this problem. Any advice?
Consider the two groups $(\mathbb{Z} \times \mathbb{Z}, \oplus)$ and $(\mathbb{Z}, +)$, where $(a,b) \oplus (c,d) = (a + c, b + d)$. Let $f: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ be a function defined by $f:(x,y) = x - y$
a) Prove that $f$ is an onto function and a homomorphism function from $(\mathbb{Z} \times \mathbb{Z}, \oplus)$ to $(\mathbb{Z}, +)$
b) Determine all $(a,b) \in \mathbb{Z} \times \mathbb{Z}$ with $f(a,b) = 0$
Would really love some help tackling this problem, thanks.

To show that $f$ is a homomorphism, we must show that $f((a,b) \oplus (c,d))=f((a,b))+f((c,d))$. So,
\begin{align}
f((a,b) \oplus (c,d)) &= f((a+c,b+d))\\
&=a+c-(b+d)\\
&=a+c-b-d\\
&=a-b+c-d\\
&=f((a,b))+f((c,d)).
\end{align}
So, $f$ is a homomorphism.
To show $f$ is onto (surjective), we must show that for every $y \in \mathbb Z$, there exists some $x \in \mathbb Z \times \mathbb Z$, such that $y=f(x)$.
We see that for $y \in \mathbb Z$, $$y=y-0=f((y,0))$$
So, $f$ is onto (surjective).
Part $b$ is asking us to find the kernel of $f$. Remember, the kernel of a homomorphism are all the elements in the domain that map to the identity element of the codomain. In this case, 
$$\text{ker } f= \{(x,y) \in \mathbb Z \times \mathbb Z \mid f((x,y))=0 \in \mathbb Z \}$$
How can we get $f((x,y))=x-y=0$? We see that if $x-y=0$, then $x=y$. So, the kernel of $f$ is $$\text{ker } f = \{(x,y) \in \mathbb Z \times \mathbb Z \mid x=y \}.$$
