= Solution
A <topological space> is <Hausdorff> when any two distinct points have <disjoint> <open neighbourhoods>.
Any two nonempty <open sets> of the infinite <cofinite topology> intersect: their <complements> are <finite sets>, so the <complement> of their <intersection> is a <finite union> of <finite sets> and cannot be all of $Z$. Thus $Z$ is a <hyperconnected space>.
If the <continuous map> $f$ took distinct values at $s,t\in Z$, choose <disjoint> <open neighbourhoods> $G,H$ of $f(s),f(t)$ in the <Hausdorff space> $Y$. By <continuity>, $f^{-1}(G)$ and $f^{-1}(H)$ would be nonempty <disjoint> <open sets> of $Z$, a contradiction. Therefore
$$
\boxed{f\text{ is constant}.}
$$
The proof establishes the general fact that <continuous maps from an infinite cofinite space to a Hausdorff space are constant>.
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