Understanding the Disjoint Union Topology

general-topology

I'm currently reading Introduction to Topological Manifolds by John Lee, and I'm trying to wrap my brain around the concept of a disjoint union topology. Here is the definition given in the book:

Suppose $(X_{\alpha})_{\alpha \in A}$ is an indexed family of nonempty topological spaces. Recall that their disjoint union is the set $\bigsqcup_{\alpha \in A} X_{\alpha}$ consisting of all ordered pairs $(x,\alpha)$ with $x \in A$ and $x \in X_{\alpha}$ (see Appendix A). For each $\alpha \in A$, there is a canonical injection $\iota_{\alpha}: X_{\alpha} \to \bigsqcup_{\alpha \in A} X_{\alpha}$ given by $\iota_{\alpha}(x) = (x,\alpha)$, and we usually identify each set $X_{\alpha}$ with its image $X_{\alpha}^{*} = \iota_{\alpha}(X_{\alpha})$.

We define the disjoint union topology on $\bigsqcup_{\alpha \in A} X_{\alpha}$ by declaring a subset of the disjoint union to be open if and only if its intersection with each set $X_{\alpha}$ (considered as a subset of the disjoint union) is open in $X_{\alpha}$. With this topology, $\bigsqcup_{\alpha \in A} X_{\alpha}$ is called a disjoint union space.

According to the definition of the disjoint union given in Appendix A,
$$ \bigsqcup_{x \in A} X_{\alpha} := \left\{(x,\alpha): x \in X_{\alpha} \text{ and } \alpha \in A \right\} $$

My questions:

  1. What does the author mean when he says "we usually identify each set $X_{\alpha}$ with its image $X_{\alpha}^{*} = \iota_{\alpha}(X_{\alpha})$? Is he saying that whenever we write "$X_{\alpha}$ we really mean $X_{\alpha}^{*}$, so that we don't want to have to always write the asterisk?

  2. What is the definition of the disjoint union topology in explicit set notation? I'm not completely sure I understand what is meant by "considered as a subset of the disjoint union". Here is my interpretation of the book's definition:

$$ \mathcal{T} := \left\{B \subset \bigsqcup_{\alpha \in A} X_{\alpha}: \pi_{1}(B) \cap X_{\alpha} \text{ is open in } X_{\alpha} \text{ for each } \alpha \in A \right\}. $$

Here, $\pi_1(B)$ denotes the image of the set $B$ under the projection map $\pi_{1}: (x,\alpha) \mapsto x$. Is my interpretation right?

  1. I'm trying to come up with some simple examples so that I can wrap my brain around this better. Is the following example correct?

Consider the spaces $\mathbb{R}$ and $\mathbb{Z}$, where $\mathbb{R}$ is equipped with the Euclidean topology and $\mathbb{Z}$ is equipped with the finite complement topology. Then if my understanding is correct, the set $B: = (-2,2) \times \{1,2\}$ is open in $\mathbb{R} \sqcup \mathbb{Z}$ because
$$ \pi_1(B) \cap \mathbb{R} = (-2,2) \text{ is open in } \mathbb{R}$$
and
$$ \pi_1(B) \cap \mathbb{Z} = \{-1,0,1\} \text{ is open in } \mathbb{Z}.$$

But the set $C := (0,\infty) \times \{1,2\}$ is not open in $\mathbb{R} \sqcup \mathbb{Z}$ because $\pi_1(C) \cap \mathbb{Z} = \{1,2,3,\ldots\}$ is not open in $\mathbb{Z}$ because neither $C$ nor $\mathbb{Z} \setminus C$ is finite.

Do I have this right? Any other insights and/or simple examples would be very much appreciated.

Best Answer

  1. That is correct. It is very convenient to think of these disjoint union spaces (coproducts) as containing the spaces they are the disjoint unions of. I can view this space as so many distinct boxes, each space living in its separate box.

  2. This is technically wrong. You use $\pi_1:\sqcup_{\alpha\in A}\to X_{\alpha}$ as if it were a valid function yet the alpha its codomain is defined for is variable. But you’re right in spirit. We describe disjoint union spaces by their inclusion maps. It is better to write: $$\mathcal{T}=\left\{B\subseteq\bigsqcup_{\alpha\in A}X_{\alpha}:i^{-1}_\alpha(B)\text{ is open for every }\alpha\in A\right\}$$This is the finest topology for which the inclusions are continuous, the so-called final/inductive/colimit/coinduced topology.

  3. Again, I know what you mean, but you shouldn’t write $\times$ in your description for $B$. Rather you should use $\sqcup$. This use of product notation rather than coproduct notation also has me quite confused with the rest of the example. Example $C$ is wrong because you use product notation, and the component $(0,\infty)$ is not actually supposed to live in $\Bbb Z$’s component. So the intersection should actually be empty. This is the reason why it is better to phrase openness in terms of $i^{-1}$ rather than this ‘projection’ $\pi$. To clarify, $C’:=M\sqcup N$ for any $M$ open in $\Bbb R$, $N$ open in $\Bbb Z$, shall be an open set.

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