The canonical injections of the disjoint union topology are topological embeddings
Contents
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:
- Continuity of [ilmath]i_\beta[/ilmath] - covered on the canonical injections of the disjoint union topology page so not shown on this pag
- [ilmath]i_\beta[/ilmath] being injective and
- [ilmath]i_\beta[/ilmath] being a homeomorphism between [ilmath]X_\beta[/ilmath] and [ilmath]i_\beta(X_\beta)[/ilmath]
The message provided is:
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] ....
- [ilmath]\overline{i_\beta}(U)\cap X_\gamma^*=\emptyset[/ilmath] and by definition, [ilmath]\emptyset\in\mathcal{J}_\gamma[/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]!
- If [ilmath]\gamma\ne\beta[/ilmath] then
- Let [ilmath]\gamma\in I[/ilmath] be given
References