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 $\{X_\alpha\}_{\alpha\in I}$.

  1. 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 $\alpha\in I$, $f_\alpha = f \circ i_\alpha$.
  2. 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.
  3. Show that each function $i_\alpha$ is an embedding, i.e., a homeomorphism onto its image.
  4. 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 $i_\alpha(X_\alpha)$.
  5. Show that $ \coprod_{\alpha\in I}X_\alpha$ is first-countable if and only if $X_\alpha$ is first-countable for each $\alpha\in I$.
  6. 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.
  7. 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.
  8. Show that if each space $X_\alpha$ is metrizable, then $ \coprod_{\alpha\in I}X_\alpha$ is metrizable.