Disjoint Unions
due by Wednesday, Nov 12, 2025
Let $\{X_\alpha\}_{\alpha\in I}$ be a collection of topological spaces. We define the Disjoint Union (or Coproduct) of the spaces as follows:
Consider the set $$\coprod_{\alpha\in I}X_\alpha = \{(\alpha, x) : \alpha\in I, x \in x_\alpha\}$$ and the collection of functions $i_\alpha : X_\alpha \to \coprod_{\alpha\in I}X_\alpha$ given by $i_\alpha(x) = (\alpha, x)\in \coprod_{\alpha\in I}X_\alpha$ for $x\in X_\alpha$. The disjoint union of the spaces $\{X_\alpha\}_{\alpha\in I}$ is the set $\coprod_{\alpha\in I}X_\alpha$ with the final topology induced by the maps $\{i_\alpha\}_{\alpha\in I}$. The disjoint union topological space is also denoted $ \coprod_{\alpha\in I}X_\alpha$.
Henceforth fix a collection of topological spaces $\{X_\alpha\}_{\alpha\in I}$.
- Show that given a topological space
$Y$and a collection of continuous functions$f_\alpha: X_\alpha \to Y$,$\alpha\in I$, there is a unique continuous function$f: \coprod_{\alpha\in I}X_\alpha \to Y$such that for each$\alpha\in I$,$f_\alpha = f \circ i_\alpha$. - State and prove a result saying that the disjoint union
$ \coprod_{\alpha\in I}X_\alpha$is unique given a universal property of the above form. - Show that each function
$i_\alpha$is an embedding, i.e., a homeomorphism onto its image. - Suppose each space
$X_\alpha$is connected and non-empty. Show that the connected components of$ \coprod_{\alpha\in I}X_\alpha$are the sets$i_\alpha(X_\alpha)$. - Show that
$ \coprod_{\alpha\in I}X_\alpha$is first-countable if and only if$X_\alpha$is first-countable for each$\alpha\in I$. - Suppose each space
$X_{\alpha}$is non-empty. Show that$ \coprod_{\alpha\in I}X_\alpha$is second countable if and only if each space$X_\alpha$is second countable and $I$ is (finite or) countable. - Suppose each space
$X_{\alpha}$is non-empty. Show that$ \coprod_{\alpha\in I}X_\alpha$is compact if and only if each space$X_\alpha$is compact and $I$ is finite. - Show that if each space
$X_\alpha$is metrizable, then$ \coprod_{\alpha\in I}X_\alpha$is metrizable.