Show that the product topology on $X\times Y$ is the same as the disjoint union topology on $\bigsqcup_{y\in Y}X$.

Question

(Exercise 3.45 Lee’s Introduction to Topological Manifolds) Let $(X,\mathscr{T})$ and $Y$ be topological spaces, the latter of which is discrete. Show that their cartesian product $X\times Y$ is equal to the disjoint union $\bigsqcup_{y\in Y}X$, and the product topology the same as the disjoint union topology.

My attempt: The first bit is easy. By definition, the two expressions are equivalent. The part the part that bothers me is the second: how is the product topology the same as the disjoint union topology? How can we show that $$\mathscr{T}^{\times} =:\bigcup_{\alpha\in A}\{U_{\alpha}\times V_{\alpha}:U_{\alpha}\in\mathscr{T}, V_{\alpha}\in \mathcal{P}(Y)\},$$ and $$\mathscr{T}^{\sqcup} =: \{U\subseteq X\times Y :U\cap X\in\mathscr{T}\},$$ are equivalent. Could anyone please give a hint which will point me in the right direction? Thanks in advance.

Answer 1

Try first showing that a basic open set in the product topology (i.e., one of the form $U\times V$ where $U\subseteq X$ and $V\subseteq Y$ are open) is open in the disjoint union topology. This doesn't require $Y$ to be discrete.

The reverse direction is more difficult. You must show that every set open in the disjoint union topology is open in the product topology. Try fixing an open subset $U$ of the disjoint union, and a point $(x_0,y_0)\in U$. Then show that there is a basic open set in the product topology $V\times W$ such that $(x_0,y_0)\in V\times W \subseteq U$. Remember that $Y$ is discrete and that, in order for $U$ to be open in the disjoint union topology, we must have that the set $\{x\in X : (x,y)\in U\}$ is open in $X$ for each $y\in Y$.

Answer 2

In general, if you have subbases $\{A_i\}$ and $\{B_j\}$ of the topologies of $X$ and $Y$ respectively, a subbasis for the product topology is given by all the $A_i \times B_j$. In this case, a subbasis for $Y$ is given by its points, so that a subbasis for the product topology is $\{A_i \times \{y\} \mid i \in I, y \in Y\}$. Can you see why this is also a subbasis for the disjoint union topology?

Answer 3

Observe that the disjoint union $\bigsqcup_{y\in Y} X$ has the universal property that it comes with maps $X \to \bigsqcup_{y\in Y} X$ for each $y\in Y$, and given a continuous map $X\to Z$ for each $y\in Y$, there exists a unique map $\bigsqcup_{y\in Y} X \to Z$ such that the compositions $X \to \bigsqcup_{y\in Y} X \to Z$ agree with the continuous maps $X\to Z$ for each $y\in Y$. Note that any two spaces that have this universal property are homeomorphic.

Now, it suffices to show that $X\times Y$ has this property. Define the maps $X\to X\times Y$ for $y\in Y$, by $X \to X\times \{y\} \to X\times Y$.

Finally, assume we have a map $f_y: X\to Z$ for each $y\in Y$. We define $X\times Y \to Z$ by $(x,y) \mapsto f_y(x)$. This is clearly the unique function such that each composition $X\to X\times Y \to Z$ agrees with $f_y$. Thus, we only need to show that the map $f: X\times Y \to Z$ is continuous. Consider an open set $U\subset Z$ and look at $f^{-1}(U) \subset X\times Y$. It should be clear that $f^{-1}(U) = \bigcup_{y\in Y} f_y^{-1}(U) \times \{y\}$, and the latter sets are all open in $X\times Y$ by definition of the product topology.

It now follows that $\bigsqcup_{y\in Y} X \cong X\times Y$.