The Stacks project

Proof. This follows from the discussion above and Categories, Lemma 4.14.11. It follows from the description above that the forgetful functor commutes with limits. Another way to see this is to use Categories, Lemma 4.24.5 and use that the forgetful functor has a left adjoint, namely the functor which assigns to a set the corresponding discrete topological space. $\square$

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

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

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

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

Proof. Follows from the description of the limit topology in Lemma 5.14.2. $\square$

Proof. Let $I$ be a set and for $i \in I$ let $X_ i$ be a quasi-compact topological space. Set $X = \prod X_ i$. Let $\mathcal{B}$ be the set of subsets of $X$ of the form $U_ i \times \prod _{i' \in I, i' \not= i} X_{i'}$ where $U_ i \subset X_ i$ is open. By construction this family is a subbasis for the topology on $X$. By Lemma 5.12.15 it suffices to show that any covering $X = \bigcup _{j \in J} B_ j$ by elements $B_ j$ of $\mathcal{B}$ has a finite refinement. We can decompose $J = \coprod J_ i$ so that if $j \in J_ i$, then $B_ j = U_ j \times \prod _{i' \not= i} X_{i'}$ with $U_ j \subset X_ i$ open. If $X_ i = \bigcup _{j \in J_ i} U_ j$, then there is a finite refinement and we conclude that $X = \bigcup _{j \in J} B_ j$ has a finite refinement. If this is not the case, then for every $i$ we can choose an point $x_ i \in X_ i$ which is not in $\bigcup _{j \in J_ i} U_ j$. But then the point $x = (x_ i)_{i \in I}$ is an element of $X$ not contained in $\bigcup _{j \in J} B_ j$, a contradiction. $\square$

Proof. Recall that $\mathop{\mathrm{lim}}\nolimits X_ i$ is a subspace of $\prod X_ i$. By Theorem 5.14.4 this product is quasi-compact. Hence it suffices to show that $\mathop{\mathrm{lim}}\nolimits X_ i$ is a closed subspace of $\prod X_ i$ (Lemma 5.12.3). If $\varphi : j \to k$ is a morphism of $\mathcal{I}$, then let $\Gamma _\varphi \subset X_ j \times X_ k$ denote the graph of the corresponding continuous map $X_ j \to X_ k$. By Lemma 5.3.2 this graph is closed. It is clear that $\mathop{\mathrm{lim}}\nolimits X_ i$ is the intersection of the closed subsets

Thus the result follows. $\square$

Proof. In the proof of Lemma 5.14.5 we have seen that $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is the intersection of the closed subsets

inside the quasi-compact space $\prod X_ i$ where $\varphi : j \to k$ is a morphism of $\mathcal{I}$ and $\Gamma _\varphi \subset X_ j \times X_ k$ is the graph of the corresponding morphism $X_ j \to X_ k$. Hence by Lemma 5.12.6 it suffices to show any finite intersection of these subsets is nonempty. Assume $\varphi _ t : j_ t \to k_ t$, $t = 1, \ldots , n$ is a finite collection of morphisms of $\mathcal{I}$. Since $\mathcal{I}$ is cofiltered, we can pick an object $j$ and a morphism $\psi _ t : j \to j_ t$ for each $t$. For each pair $t, t'$ such that either (a) $j_ t = j_{t'}$, or (b) $j_ t = k_{t'}$, or (c) $k_ t = k_{t'}$ we obtain two morphisms $j \to l$ with $l = j_ t$ in case (a), (b) or $l = k_ t$ in case (c). Because $\mathcal{I}$ is cofiltered and since there are finitely many pairs $(t, t')$ we may choose a map $j' \to j$ which equalizes these two morphisms for all such pairs $(t, t')$. Pick an element $x \in X_{j'}$ and for each $t$ let $x_{j_ t}$, resp. $x_{k_ t}$ be the image of $x$ under the morphism $X_{j'} \to X_ j \to X_{j_ t}$, resp. $X_{j'} \to X_ j \to X_{j_ t} \to X_{k_ t}$. For any index $l \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ which is not equal to $j_ t$ or $k_ t$ for some $t$ we pick an arbitrary element $x_ l \in X_ l$ (using the axiom of choice). Then $(x_ i)_{i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})}$ is in the intersection

by construction and the proof is complete. $\square$