# Gluing topological spaces

The following is called a gluing datum:

• A family of topological spaces $(U_i)$,
• For all $i,j$ an open subset $U_{ij}\subseteq U_i$
• For all $i,j$ a map $\varphi_{ji}:U_{ij}\to U_{ji}$

such that

• $\varphi_{ii}=\operatorname{id}$ for aall $i$,
• $\varphi_{ji}(U_{ij}\cap U_{ik})\subseteq U_{ji}\cap U_{jk}$ for all $i,j,k$,
• $\varphi_{kj}\circ \varphi_{ji}=\varphi_{ki}$ on $U_{ij}\cap U_{ik}$ for all $i,j,k$.

Then there is a space $X$ together with open continous functions $\psi_i:U_i\to X$, such that

1) $X$ is covered by the images of the $U_i$,

2) $\psi_j\circ\varphi_{ji}=\psi_i$ on $U_{ij}$ for all $i,j$,

3) $\psi_i(U_i)\cap\psi_j(U_j)=\psi_i(U_{ij})=\psi_j(U_{ij})$ for all $i,j$.

Moreover, $X$ is universal with respect to 2).

Question: What is happening here, categorically? Is this a special colimit? Does this have a name in category theory? Which parts are categorical nonsense and which are topology?

Thank you.

• I cannot address the category theory, but I can tell you what the topology is. Generally speaking, when you have a bare, unadorned set $X$, a collection of topological spaces $U_i$, and a collection of functions $f_i : U_i \to X$, then there is an induced topology on $X$ which is the intersection of all topologies $\mathcal{T}$ such that each of the functions $f_i$ is continuous with respect to $\mathcal{T}$. One says this is the "weakest" or "coursest" topology with respect to which each $f_i$ is continuous. – Lee Mosher Sep 12 '16 at 17:10
• Hint: To show that it is a colimit, you have to prove that $1)$ and $3)$ follow from the assumptions and $2)$, but for this, you have to make sure to take the right 'meaning' of universal with respect to $2)$. – Ben Sep 12 '16 at 17:16
• A related, dual, (but not quite the same) categorial notion: math.stackexchange.com/q/1619737/12885 – Ben Sep 12 '16 at 17:51
• @LeeMosher But here, a priori there is no set $X$ and no functions $f_i\to U$, which are used to equip $X$ with a topology, right? $X$ and $\psi_i$ are constructed from the gluing data. – user114885 Sep 13 '16 at 9:11
• Good point. Again, I don't know the category theory words, but there is a straightforward description of the set $X$: it is the quotient set of the disjoint union of the sets $U_i$, using the equivalence relation that is generated by the relation where $x \in U_i$ is related to $y \in U_j$ if there exists $j$ such that $x \in U_{ij}$, $y \in U_{ji}$, and $\phi_{ji}(x)=y$. – Lee Mosher Sep 13 '16 at 12:34

I will assume that gluing data is also meant to include the condition $U_{ii} = U_i$.

I'm too tired to organize this all in a narrative, so this will be fairly disjointed.

It is fairly common in category theory to consider two families of objects $X_k$ and $X_{ij}$, families of maps $f_{ij} : X_{ij} \to X_i$ and $g_{ij} : X_{ij} \to X_j$, and the corresponding coequalizer

$$\coprod_{ij} X_{ij} \overset{f}{\underset{g}{\rightrightarrows}} \coprod_k X_k \xrightarrow{\rho} X$$

or, if we include 'redundant' $X_{ii} = X_i$, a pushout square

$$\begin{matrix} \coprod_{ij} X_{ij} &\xrightarrow{f}& \coprod_k X_k \\ \!g\!\downarrow & & \downarrow\!\rho\! \\ \coprod_k X_k &\xrightarrow{\rho}& X \end{matrix}$$

The picture here is that the $X_k$'s are a description of an object of interest, and the $X_{ij}$'s describe relations between the descriptions.

Geometrically, we might think of the $X_k$'s as a cover, and the $X_{ij}$'s describe the overlap between them. Algebraically, we might think of $X_k$ as being generators, and the $X_{ij}$ as being relations.

Either way, it's clear that in the "nicest" arrangement, we want each $X_{ij}$ to to be the pullback of $X_i \to X \leftarrow X_{j}$, so that it truly does describe all relations between $X_i$ and $X_j$, and you only really want one $X_{ij}$ per pair of indices.

Gluing data is in this nice situation, but not all such diagrams are: e.g. we might only have a more minimal description of the relations, or the covers might actually self-intersect nontrivially (e.g. taking the interval $X_1 = [0,2\pi]$ as a cover of the circle, with $X_{11}$ being a single point mapped to both ends) or other deficiencies may apply.

The gluing data gives an example of this sort of diagram: on each $U_{ij}$, the two maps into $\coprod_k U_k$ come from

• $U_{ij} \hookrightarrow U_i$
• $U_{ij} \xrightarrow{\varphi_{ji}} U_{ji} \hookrightarrow U_j$

Furthermore, the data does assert there is "just one" in the sense that the $\varphi_{ji}$ gives a homeomorphism $U_{ij} \to U_{ji}$, and they do so coherently in the sense that $\varphi_{ij} = \varphi_{ji}^{-1}$ and $\varphi_{ii} = 1_{U_i}$, so different "paths" $U_{ij} \to U_{ji}$ (e.g $\varphi_{ji}$ versus $\varphi_{ji} \circ \varphi_{ii} \circ \varphi_{ij} \circ \varphi_{ji}$) all give the same map.

The property that Top has that makes this setup convenient to work with is that it is infinitary extensive — i.e. that coproducts really do act like disjoint unions. I do not know if this is actually required, but all the ways I want to reason about gluing data rely on it.

Sometimes, we also want to consider another family of $X_{ijk}$, this time with three maps down to the various $X_{mn}$. We have this in gluing data too: we can define $U_{ijk} = U_{ij} \cap U_{ik}$. And again we have "just one" per triple of indices, because we again have coherent homeomorphisms between the different permutations. It is enough to check

• $U_{ijk} = U_{ikj}$
• $\varphi_{ji} : U_{ijk} \to U_{jik}$ is a homeomorphism
• The two homeomorphism $U_{ijk} \to U_{kji}$ given by $\varphi_{ki}$ and $\varphi_{kj} \circ \varphi_{ji}$ are the same.

We could go further. This leads to a simple example of a simplicial object.

Top is nice enough to talk about relations. The colimit defining $X$ can be viewed as taking the quotient of $\coprod_k U_k$ by the relation that the two maps $\coprod_{ij} U_{ij} \rightrightarrows \coprod_k U_k$ give equivalent outputs for each input.

Normally, this relation is not an equivalence relation, and so the quotient is by the equivalence relation generated by this relation.

However, the neat thing about having the transition maps $\varphi$ is that the relation really is an equivalence relation, so the colimit is much, much easier to work with. The interesting part is that it is a transitive relation, which you can check by noting that

$$x \sim \varphi_{ji}(x) \quad \text{and} \quad \varphi_{ji}(x) \sim \varphi_{kj}(\varphi_{ji}(x))$$

only makes sense when $x \in U_{ijk}$, and that the transitive property requires $x \sim \varphi_{ki}(x)$, which we have.

In Top, having the transition maps implies that the $\psi_i : U_i \to X$ are monic. I think you can even argue it's regular monic. Off hand I don't know what you want from the category to say such things.

Finally, last feature is about open subspaces. While seemingly the part most topological in flavor, it too has an abstract analog.

Top has a open subspace classifier. Let $S = \{0,1\}$ be the Sierpinski space, with topology $\{\varnothing, \{1\}, S\}$. Then there is a natural bijection between open subspaces of $X$ and continuous maps $X \to S$:

• For every open subset $U \subseteq X$, the characteristic function $\chi_U : X \to S$ is continuous
• For any continuous map $\chi : X \to S$, the inverse image $f^{-1}(1)$ is an open subset

You can thus show that $\psi_i(U_i) \subseteq X$ is an open subspace by showing that the maps $$\chi_{ij} : \coprod_i U_j \to S : x \mapsto \begin{cases} 1 & x \in U_{ji} \\ 0 & x \notin U_{ji} \end{cases}$$ induce a well-defined map $\chi_i : X \to S$, and that $\chi_i^{-1}(1) = \psi_i(U_i)$.

Off hand, I'm not sure what additional properties you need (if any) to be able to make this argument in Top.

Here comes a diagram in $\mathrm(Top)$ whose colimit is the desired glueing. Given a glueing datum as in the question text, the diagram will consist of

• all $U_i$ and all $U_{ij}$ as objects and
• the inclusion maps $U_{ij}\to U_i$ as well as all $\varphi_{ij}$ as morphisms.

This feels a bit ad-hoc, I have to admit, but it works. If we were to write down a more general/abstract diagram, it were harder to translate that the $U_{ij}$ are actually subspaces of the $U_i$ and not just merely spaces mapping to $U_i$, but besides that, it's not too hard. But under weaker assumptions, I think $3)$ might not hold for the colimit.

Let me recall (from Lee Mosher's comment) that the following space, with the obvious maps, is a colimit for the above diagram: $$X := \coprod\nolimits_i U_i\,/{\sim},\text{ where }\sim\text{ identifies }U_{ij}\text{ with }U_{ji}\text{ via }\varphi_{ij} = \varphi_{ji}^{-1}.$$ This space is easily seen to fulfil the properties $1), 2), 3)$, so actually, there is nothing more to prove, by uniqueness. Nevertheless, we should look at what's going on in case there are only two spaces $U_1,U_2$ with subspaces $U_{12}\subset U_1$, $U_{21}\subset U_2$, with continuous maps $\varphi_{12}\colon U_{21}\to U_{12}$ and $\varphi_{21}\colon U_{12}\to U_{21}$, being mutually inverse. Then the diagram is just $$\newcommand{\ra}{\!\!\!\!\!\!\!\!\!\!\!\!\mathop{\rightleftarrows}\limits^{#1}_{#2}\quad\!\!\!\!\!\!\!\!} \newcommand{\da}{\left\downarrow{\scriptstyle#1}\vphantom{\int}\right.} % \begin{array}{rcl} U_{12} & \ra{\varphi_{21}}{\varphi_{12}} & U_{21} \\ \da{} & & \da{} \\ U_{1} & & U_2 \\ \end{array}$$ A colimit of this diagram boils down to a space $X$ with maps $\psi_i\colon U_i\to X$ such that the following diagram commutes, and the universal property that the continuous maps $f\colon X\to Y$ are, via pull-back to $U_1$ and $U_2$, in bijection with pairs $f_i = f\circ\psi_i\colon U_i\to Y$ such that $(f_1)|_{U_{12}} = f_2\circ\varphi_{21}$ (and then automatically also $(f_2)|_{U_{21}} = f_1\circ\varphi_{12}$).

$$\newcommand{\ra}{\!\!\!\!\!\!\!\!\!\!\!\!\mathop{\rightleftarrows}\limits^{#1}_{#2}\quad\!\!\!\!\!\!\!\!} \newcommand{\da}{\left\downarrow{\scriptstyle#1}\vphantom{\int}\right.} % \begin{array}{rcl} U_{12} & \ra{\varphi_{21}}{\varphi_{12}} & U_{21} \\ \da{} && \da{} \\ U_{1} & & U_2 \\ &\searrow^{\psi_1}\;\;\;\;\swarrow_{\psi_2}&\\ & X & \\ \end{array}$$

We must show that this implies $1)$ and $3)$ as in the question. This is where we see what's formal and what's rather special to $(Top)$. The universal property implies that $\coprod_i U_i\xrightarrow{\psi} X$ is an epimorphism and this is enough for the map so be surjective (in $\mathrm{(Top)}$!); therefore, $X$ is covered by $U_1$ and $U_2$ and $1)$ holds. $3)$, however, is more involved. It relies on the fact that (in $\mathrm{(Top)}$!) some push-out squares are also pull-back squares. I'm running out of time for more details right now, so let's just say that $3)$ holds relies very much on the fact that we're talking about $\mathrm{(Top)}$ and on the assumption that the $U_{ij}\subset U_i$ be subspaces. (I think I will only provide more details upon request.)