If you have a set $X$ which is equipped with a topology: what is required for the function $f: X\to \mathbb Z$ to be continuous?
The fact that $X$ has a topology should certainly be helpful. But $\mathbb Z$ is a discrete space, so I am not totally sure how to establish continuity.
I need to establish that every open set $A\subset \mathbb Z$ has an open pre-image $f^{-1}(A)$ in $X$. But what are open sets in $\mathbb Z$, and how could they be linked to the topology in $X$? Does the topology on $X$ imply, via $f$, a topology on $f(X)\subset\mathbb Z$, and hence (somehow) on all of $\mathbb Z$?
Every subset of $\Bbb Z$ is open, so the preimage of every subset of $\Bbb Z$ must be open in $X$ in order for $f$ to be continuous. In particular, let $U_n=f^{-1}[\{n\}]$ for each $n\in\Bbb Z$: in order for $f$ to be continuous, it is necessary and sufficient that all of the sets $U_n$ be open. Of course, since each $U_n=X\setminus\bigcup_{k\in\Bbb Z\setminus\{n\}}U_k$, each $U_n$ is also closed in $X$. In other words, $X$ must have a partition into countably many clopen (i.e., both closed and open) subsets, and $f$ must send each of these clopen sets to an integer.
A non-trivial example is given by letting $X=\Bbb Z\times\Bbb R$ with the product topology and letting $f$ be the projection of $X$ to the first factor:
$$f:X\to\Bbb Z:\langle n,x\rangle\mapsto n\;.$$
Then $f$ is continuous, because each of the sets $\{n\}\times\Bbb R$ is clopen in $X$.