**Proof.**
The construction of the limit given above shows that $X \subset \prod X_ i$ with the induced topology. A basis for the topology of $\prod X_ i$ are the opens $\prod U_ i$ where $U_ i \subset X_ i$ is open and $U_ i = X_ i$ for almost all $i$. Say $i_1, \ldots , i_ n \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ are the objects such that $U_{i_ j} \not= X_{i_ j}$. Then

\[ X \cap \prod U_ i = f_{i_1}^{-1}(U_{i_1}) \cap \ldots \cap f_{i_ n}^{-1}(U_{i_ n}) \]

For a general limit of topological spaces these form a basis for the topology on $X$. However, if $\mathcal{I}$ is cofiltered as in the statement of the lemma, then we can pick a $j \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and morphisms $j \to i_ l$, $l = 1, \ldots , n$. Let

\[ U_ j = (X_ j \to X_{i_1})^{-1}(U_{i_1}) \cap \ldots \cap (X_ j \to X_{i_ n})^{-1}(U_{i_ n}) \]

Then it is clear that $X \cap \prod U_ i = f_ j^{-1}(U_ j)$. Thus for any open $W$ of $X$ there is a set $A$ and a map $\alpha : A \to \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and opens $U_ a \subset X_{\alpha (a)}$ such that $W = \bigcup f_{\alpha (a)}^{-1}(U_ a)$. Set $J = \mathop{\mathrm{Im}}(\alpha )$ and for $j \in J$ set $U_ j = \bigcup _{\alpha (a) = j} U_ a$ to see that $W = \bigcup _{j \in J} f_ j^{-1}(U_ j)$. This proves (1).

To see (2) suppose that $\bigcup _{j \in J} f_ j^{-1}(U_ j)$ is quasi-compact. Then it is equal to $f_{j_1}^{-1}(U_{j_1}) \cup \ldots \cup f_{j_ m}^{-1}(U_{j_ m})$ for some $j_1, \ldots , j_ m \in J$. Since $\mathcal{I}$ is cofiltered, we can pick a $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and morphisms $i \to j_ l$, $l = 1, \ldots , m$. Let

\[ U_ i = (X_ i \to X_{j_1})^{-1}(U_{j_1}) \cup \ldots \cup (X_ i \to X_{j_ m})^{-1}(U_{j_ m}) \]

Then our open equals $f_ i^{-1}(U_ i)$ as desired.
$\square$

## Comments (2)

Comment #656 by Wei Xu on

Comment #666 by Johan on