Example 7.21.9. Let $X$ be a topological space. Let $i : Z \to X$ be the inclusion of a subset (with induced topology). Consider the functor $u : X_{Zar} \to Z_{Zar}$, $U \mapsto u(U) = Z \cap U$. At first glance it may appear that this functor is cocontinuous as well. After all, since $Z$ has the induced topology, shouldn't any covering of $U\cap Z$ it come from a covering of $U$ in $X$? Not so! Namely, what if $U \cap Z = \emptyset $? In that case, the empty covering is a covering of $U \cap Z$, and the empty covering can only be refined by the empty covering. Thus we conclude that $u$ cocontinuous $\Rightarrow $ every nonempty open $U$ of $X$ has nonempty intersection with $Z$. But this is not sufficient. For example, if $X = \mathbf{R}$ the real number line with the usual topology, and $Z = \mathbf{R} \setminus \{ 0\} $, then there is an open covering of $Z$, namely $Z = \{ x < 0\} \cup \bigcup _ n \{ 1/n < x\} $ which cannot be refined by the restriction of any open covering of $X$.

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