# The canonical injections of the disjoint union topology are topological embeddings

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Important for progress!

## Statement

Let [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be a collection of topological spaces and let [ilmath](\coprod_{\alpha\in I}X_\alpha,\mathcal{J})[/ilmath] be the disjoint union space of that family. With this construction we get some canonical injections:

• For each [ilmath]\beta\in I[/ilmath] we get a map (called a canonical injection) [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] given by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath]

We claim that each [ilmath]i_\beta[/ilmath] is a topological embedding[1] (that means [ilmath]i_\beta[/ilmath] is injective and continuous and a homeomorphism between [ilmath]X_\beta[/ilmath] and [ilmath]i_\beta(X_\beta)[/ilmath] (its image))

## Proof

Let [ilmath]\beta\in I[/ilmath] be given.

• The proof that [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath] consists of three parts:
1. Continuity of [ilmath]i_\beta[/ilmath] - covered on the canonical injections of the disjoint union topology page so not shown on this pag
2. [ilmath]i_\beta[/ilmath] being injective and
3. [ilmath]i_\beta[/ilmath] being a homeomorphism between [ilmath]X_\beta[/ilmath] and [ilmath]i_\beta(X_\beta)[/ilmath]
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Done on paper, it isn't hard. I want to save my work now though NOTES ARE BELOW

I only cover part 3 here.

We have shown [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] is continuous and injective. It only remains to show that it is a homeomorphism onto (in the surjective sense of the word "onto") its image.

• First note that every injection yields a bijection onto its image
• Thus we get a map [ilmath]\overline{i_\beta}:X_\beta\rightarrow i_\beta(\coprod_{\alpha\in I}X_\alpha)[/ilmath] given by [ilmath]\overline{i_\beta}:x\mapsto i_\beta(x)[/ilmath] (note that this means [ilmath]\overline{i_\beta}:x\mapsto(\beta,x)[/ilmath]) which is a bijection
• Next note that every bijection yields an inverse function, so now we have [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath]
• We only really need to show that [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath] is continuous
• Let [ilmath]U\in\mathcal{J}_\beta[/ilmath] (so [ilmath]U[/ilmath] is open in [ilmath]X_\beta[/ilmath]) be given
• Then we must show that [ilmath]((\overline{i_\beta})^{-1})^{-1}(U)\in\mathcal{J} [/ilmath] in order for [ilmath](\overline{i_\beta})^{-1}:i_\beta(\coprod_{\alpha\in I}X_\alpha)\rightarrow X_\beta[/ilmath] to be continuous
• But! [ilmath]((\overline{i_\beta})^{-1})^{-1}(U)=\overline{i_\beta}(U)[/ilmath]
• Recall we defined a set to be open in [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath] if its intersection with (the image of) each [ilmath]X_\alpha[/ilmath] is open in [ilmath]X_\alpha[/ilmath] Caution:Terminology is a bit fuzzy here. I need to fix that
• Let [ilmath]\gamma\in I[/ilmath] be given
• If [ilmath]\gamma\ne\beta[/ilmath] then
• [ilmath]\overline{i_\beta}(U)\cap X_\gamma^*=\emptyset[/ilmath] and by definition, [ilmath]\emptyset\in\mathcal{J}_\gamma[/ilmath]
• Caution:This is where the notation gets weird. The image of the emptyset is the empty set, not sets of the form [ilmath](\beta,x)[/ilmath] ....
• If [ilmath]\gamma=\beta[/ilmath] then
• [ilmath]\overline{i_\beta}(U)\cap X_\beta^*=U[/ilmath] Caution:or the image of [ilmath]U[/ilmath] - which is open as [ilmath]U\in\mathcal{J}_\beta[/ilmath]!