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.
Best Answer
Hint: There is an easier way, if $\pi_i:X_1 \times X_2 \rightarrow X_i $ and $p_i:X'_1 \times X'_2 \rightarrow X'_i $ are the projections and $f_i: X_i \rightarrow X'_i$ are continuous for $i=1,2$, then simply show that $p'_i\circ (f\times g) = f_i \circ \pi_i$ which is continuous by assumption.