What exactly is a topological sum?

Question

I am reading Dieck's Algebraic Topology and the following definition is given:

Let $(X_j : j\in J)$ be a family of non-empty pairwise disjoint spaces. The set $$\mathcal O = \{U\subset \coprod X_j : U\cap X_j\subset X_j \text{ open for all } j\} $$ is a topology on the disjoint union $\coprod X_j$. We call $(\coprod X_j, \mathcal O)$ the topological sum of the $X_j$.

I am having trouble understanding this definition. I understand that for an arbitrary family of sets $(A_j : j\in J)$ the disjoint union is the set $$ \coprod_{j\in J} A_j = \bigcup_{j\in J}\{(x,j):x\in A_j\}, $$ so that even if for $i\ne j$ with $A_i$ and $A_j$ not disjoint, the sets $A_i^*$ and $A_j^*$ are disjoint, where $A_i^* = \{(x,i):x\in A_i\}$. However, I do not see the following:

1. Why are the $X_j$ required to be pairwise disjoint in this construction, if the notion of disjoint union is defined even if they are not? Is it so that $\mathcal O$ is in fact a topology?
2. What does an element of $\mathcal O$ look like? To keep things simple, let's take $J=\{1,2\}$ so we are considering $X_1+X_2$ (the "sum" notation). The index notation is confusing me, so I don't know how you would write an open set explicitly.
3. What is the motivation for calling this a "sum"? It looks more like a product to me. In fact the command for $\coprod$ is \coprod. So should I think of this as a coproduct, or "categorical sum"? I have not studied any category theory, so this is not familiar to me.

Answer 1

With regard to your third question, while I agree with the comments that it's neither necessary nor beneficial to study category theory as a precursor to learning topology, I would make the case for picking up a little category theory along the way, as it becomes relevant (the "just-in-time category theory" approach). For example, when you learn about the product topology and the topological sum, it's appropriate to devote some effort to understanding the universal properties of the product and the coproduct, and when you start learning about the fundamental group or homology, you'll want to learn what a functor is. It may not be strictly necessary in order to understand the topology, but let me give two reasons why picking up this small amount of category theory may be beneficial, in the particular case of products and coproducts.

1. It motivates the construction of the topological product and the topological sum. Sure, you can construct this topology on the disjoint union of some collection of spaces, making it into the topological sum, but why should you do such a thing? Why is the topological sum a thing worth considering? There are many possible answers, but one of them is that the topological sum is the coproduct in the category of topological spaces and continuous functions. Without even worrying about what that means precisely, it says that the topological sum behaves analogously to coproducts in other categories. So, if you believe that the disjoint union of sets (which is the coproduct in the category of sets) or the direct sum of groups or vector spaces (which are the coproduct in the categories of groups or vector spaces, respectively) are interesting or useful, then probably the topological sum is also interesting or useful.

An even better example of this idea is the product of topological spaces. You form the product of spaces by taking the cartesian product of their underlying sets, and then putting a suitable topology on it. Well, there are (at least) two topologies a reasonable person might try, the product topology and the box topology, which are different if you're taking the product of infinitely many spaces. I don't know about you, but when I learned about these for the first time, I was convinced that the box topology was a more sensible choice (why should we restrict all but finitely many of the open sets to be the entire space??). Well, the product topology is the product in the category of topological spaces, and the box topology isn't.

2. More practically, the universal property is useful for producing continuous maps. Have you ever wanted to define a continuous map from the topological sum to some other space? (If you haven't, you will.) More concretely, let's say you want to build a continuous map $$ \coprod_{i \in I} X_i \to Z $$ where the $X_i$'s and $Z$ are some topological spaces. It turns out that the topological sum has exactly the right topology to make this an easy task. All you have to do is produce a continuous map $$ X_i \to Z $$ for each $i$! The universal property of the coproduct says that choosing continuous maps $X_i \to Z$ uniquely determines a continuous map $\coprod X_i \to Z$. I'll grant that you can understand and use this universal property without stating it in the language of category theory, but the same universal property applies to coproducts in any category, so phrasing it in categorical language makes clear similarity in the behavior of objects (such as topological sums and direct sums of vector spaces) that otherwise look completely different.

Answer 2

Just take two copies of $\mathbb{R}$ for definiteness, so $\Bbb R + \Bbb R$ in sum notation (one also sees $\Bbb R \oplus \Bbb R$ sometimes).

So as $\mathbb{R}$ is not disjoint from itself, we have to make them disjoint by using "labels", so we can tell for a point in the whole sum/union from which summand it came from; the usual way that you describe is to form pairs and take a union of those, as a set then $\mathbb{R} + \mathbb{R}$ equals $$\{(x,i): x \in \mathbb{R}, i \in \{0,1\}\}$$ so a point always comes with a label $0$ or $1$, uniquely determined.

The topology we put on it is just two copies of the the topology of $\mathbb{R}$, one for each summand and just take unions of them. So an open set is of the form $(O_1 \times \{0\}) \cup (O_2 \times \{1\})$, where $O_1$ is any open subset of $\mathbb{R}$ and so is $O_2$ of $\mathbb{R}$ (in general we'd have spaces $X_1$ and $X_2$ respectively that we take the open sets from), so they've got a unique "open part" in each summand (which could also be empty, so just taking $(-1,1)$ in the left copy, so $(-1,1)\times \{0\}$ is also a valid choice. And if $A$ is not open in $\mathbb{R}$, $A \times \{0\}$ will not be open in $\mathbb{R} + \mathbb{R}$ as well. So open sets are pretty simple: both parts in both summands should be open in their original space.

If you want to be formal, we can define $j_0: \mathbb{R} \to \mathbb{R} + \mathbb{R}$ and $j_1: \mathbb{R} \to \mathbb{R}+ \mathbb{R}$ by $j_0(x)=(x,0)$ and $j_1(x)=(x,1)$ and define the topology as

$$\{O \subseteq \mathbb{R} + \mathbb{R} \mid j_0^{-1}[O] \text{ open in } \mathbb{R} \text{ and } j_1^{-1}[O] \text{ open in } \mathbb{R}\}$$

which can be checked to be the strongest topology that on $\mathbb{R} + \mathbb{R}$ that makes both $j_0$ and $j_1$ continuous.

See my post here for some more general considerations on such so-called final topologies.

So in a way, you just take (in this case two) independent copies of the spaces we are "summing", and put an obvious topology on it that ensures both summands are natural subspaces of the sum space (via the (open) embeddings $j_0,j_1$ here). The copies are totally separated from each other: each copy is clopen in the sum, so we almost always get disconnected spaces. And infinite sums (i.e. infinitely many summands) are rarely compact, as we get an open cover of summand-copies that we cannot reduce. They're loose and independent pieces. Maybe that's why they're often combined with quotients to glue parts together again, via another final topology construction. They can be a handy technical construct but normal products are (IMHO) much more important, and preserve more properties.