Disjoint union topology vs Product topology

Question

I am trying to understand the differens between the two more precisely. I am aware that the product topology is the product in category while the disjoint union topology is the coproduct. Unfortunately I feel that doesn't quite explain it. If we let $(\mathbf{A},\alpha)$ and $(\mathbf{B},\beta)$ be two given topologies. I understand that our product topology for $\mathbf{A}\times\mathbf{B}$ is that a set $(u,v)$ is open in it if and only if $u$ is open in $\mathbf{A}$ and $v$ is open in $\mathbf{B}$. So far so good, this is all crystal clear to me.

I will use the $\oplus$ symbol rather than the upside-down product sign, I don't like it. Let our disjoint union $\mathbf{A}\oplus\mathbf{B}$, a set for us here is then $u\oplus v$ and this is where I am starting to feel uncertain. I get there is a very evident inclusion map $\imath_A:\mathbf{A}\to\mathbf{A}\oplus\mathbf{B}$ which would be $\imath_A(a)=a\oplus\emptyset$, or more I deduce this from the context of what I am reading that this is the most likely candidate. I also gather from what I have read (I will inform I have not found it in a proper book in my studies so far, might have missed it but I have read this primarely online from definitions) that a set $u\oplus v$ is open if and only if $\imath_A^{-1}(u\oplus v)$ and $\imath_B^{-1}(u\oplus v)$ are open.

This doesn't strike me as really that different, or different at all, from the product topology at first sight. I will speculate a bit here so please tell me what I get right and what I get wrong. I know the product topology is the cartesian product hence no part of the pair $(a,b)$ can be empty, the empty set is of course always part in the topology by definition. What I gather from my reading though is that they often place emphasis on the word "or", as in for $w\in\mathbf{A}\oplus\mathbf{B}$ we have that either $w\in\mathbf{A}$ OR $w\in\mathbf{B}$. This leads me to conclude that if we have $a\oplus b\in\mathbf{A}\oplus\mathbf{B}$ then either $a=\emptyset$ or we have that $b=\emptyset$. In a more word friendly way, that any element in our disjoint union topology is a member of either of the components that makes up the union, and not a construct as a combination from both of the components, which it is in a proudct topology.

Am I understanding this correctly or am I missing something?

Answer 1

Not quite. First a minor point that may or may not help to solve some of your confusion. If $a$ is a subset of $A$ and $b$ is a subset of $B$, then it does not make sense to write $a \oplus b \in A \oplus B$, but $a \oplus b \subseteq A \oplus B$ would make sense; $a \oplus b$ is not an element of $A \oplus B$ but a subset thereof. Similarly, writing $i_A(a) = a \oplus \emptyset$ is problematic, since $a \oplus \emptyset$ is not an element of $A \oplus B$. Moreover, it need not be the case that one of $a$ or $b$ is necessarily empty. For instance, if $A = \{1,2,3\}$ and $B = \{4,5\}$, then $\{1, 5\} \subseteq A \oplus B$ but both $i_A^{-1}(\{1,5\}) = \{1\}$ and $i_B^{-1}(\{1,5\}) = \{5\}$ are non-empty.

Note that this really has nothing to do with the topologies; here we are only thinking about what $A \oplus B$ is like as a set. Indeed, figuring out the difference between $A \oplus B$ and $A \times B$ as sets would be a good starting point: if they are different sets, then certainly they will also be different topological spaces.

Take $A = \{1, 2, 3\}$ and $B = \{4, 5\}$ as above. Then $A \oplus B = \{1, 2, 3, 4, 5\}$ consists of $5$ points (but depending on your treatment of disjoint unions, you may want to write this as $A \oplus B = \{(1,0), (2,0), (3,0), (4,1), (5,1)\}$ or something like that). In any case, the set will consist of $5$ elements.)

On the other hand, $A \times B = \{(1,4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)\}$ consists of $6$ points so the two sets are really different (in that they have different cardinalities and there can be no bijection (thus in particular no homeomorphism) between them).

However, coming to the actual topologies, I can see where your confusion might arise. In both the case of the disjoint union topology and the product topology we have very natural maps: in the first case we have the inclusions, and in the second case we have the projections. When we cook up the two topologies the way we do, it ensures that in the first case, the inclusions become continuous for free, and in the second case, the projections become continuous for free, so certainly there are analogues between the two notions. You already seem to be aware of this though, as you mention products vs. coproducts.

Answer 2

The difference between the two topologies all lies in their canonical projections.

Let $\{X_i:i\in I\}$ be a family of topological spaces. Consider $X=\bigsqcup{X_i}$ . Now consider the canonical injections for each $i\in I$ given by the map $f_i: X_i\to X$ given by $f_i(x)=(x,i)$. The disjoint union topology is the topology in which all the $f_i$'s are continuous.

Now the product topology: Consider $X= \prod{X_i}$. Now consider the canonical projections $p_{i}X \to X_{i}$. The product topology is that topology in which all the $p_i$'s are continuous.

The difference here lies in the canonical mappings which happen to define a topology for the product of $X_i$ and the disjoint union of $X_i$.

Also you will see that there are elements in the disjoint union topology which are open in one of the constituent unions. That is if $X= M\sqcup N$ is given the disjoint union topology then there are open set $P$ in $X$ which is open in $M$ and sets $Q$ in $X$ which are open in $N$. But in the product topology an open set has to open in both the sets.