## 5.14 Limits of spaces

The category of topological spaces has products. Namely, if $I$ is a set and for $i \in I$ we are given a topological space $X_ i$ then we endow $\prod _{i \in I} X_ i$ with the product topology. As a basis for the topology we use sets of the form $\prod U_ i$ where $U_ i \subset X_ i$ is open and $U_ i = X_ i$ for almost all $i$.

The category of topological spaces has equalizers. Namely, if $a, b : X \to Y$ are morphisms of topological spaces, then the equalizer of $a$ and $b$ is the subset $\{ x \in X \mid a(x) = b(x)\} \subset X$ endowed with the induced topology.

Lemma 5.14.1. The category of topological spaces has limits and the forgetful functor to sets commutes with them.

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. $\square$

Lemma 5.14.2. Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_ i$ be a diagram of topological spaces over $\mathcal{I}$. Let $X = \mathop{\mathrm{lim}}\nolimits X_ i$ be the limit with projection maps $f_ i : X \to X_ i$.

1. Any open of $X$ is of the form $\bigcup _{j \in J} f_ j^{-1}(U_ j)$ for some subset $J \subset I$ and opens $U_ j \subset X_ j$.

2. Any quasi-compact open of $X$ is of the form $f_ i^{-1}(U_ i)$ for some $i$ and some $U_ i \subset X_ i$ open.

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. $\square$

Lemma 5.14.3. Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_ i$ be a diagram of topological spaces over $\mathcal{I}$. Let $X$ be a topological space such that

1. $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a set (denote $f_ i$ the projection maps),

2. the sets $f_ i^{-1}(U_ i)$ for $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and $U_ i \subset X_ i$ open form a basis for the topology of $X$.

Then $X$ is the limit of the $X_ i$ as a topological space.

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. $\square$

The following lemma does not hold if one drops the assumption that the spaces $X_ i$ are Hausdorff, see Examples, Section 109.4.

Lemma 5.14.5. Let $\mathcal{I}$ be a category and let $i \mapsto X_ i$ be a diagram over $\mathcal{I}$ in the category of topological spaces. If each $X_ i$ is quasi-compact and Hausdorff, then $\mathop{\mathrm{lim}}\nolimits X_ i$ is quasi-compact.

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

$\Gamma _\varphi \times \prod \nolimits _{l \not= j, k} X_ l \subset \prod X_ i$

Thus the result follows. $\square$

The following lemma generalizes Categories, Lemma 4.21.7 and partially generalizes Lemma 5.12.6.

Lemma 5.14.6. Let $\mathcal{I}$ be a cofiltered category and let $i \mapsto X_ i$ be a diagram over $\mathcal{I}$ in the category of topological spaces. If each $X_ i$ is quasi-compact, Hausdorff, and nonempty, then $\mathop{\mathrm{lim}}\nolimits X_ i$ is nonempty.

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

$Z_\varphi = \Gamma _\varphi \times \prod \nolimits _{l \not= j, k} X_ l$

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

$Z_{\varphi _1} \cap \ldots \cap Z_{\varphi _ n}$

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

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