The canonical injections of the disjoint union topology are topological embeddings

Stub grade: A

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Important for progress!

Statement

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

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

We claim that each $i_\beta$ is a topological embedding (that means $i_\beta$ is injective and continuous and a homeomorphism between $X_\beta$ and $i_\beta(X_\beta)$ (its image))

Proof

Let $\beta\in I$ be given.

• The proof that $i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha$ by $i_\beta:x\mapsto(\beta,x)$ consists of three parts:
  1. Continuity of $i_\beta$ - covered on the canonical injections of the disjoint union topology page so not shown on this pag
  2. $i_\beta$ being injective and
  3. $i_\beta$ being a homeomorphism between $X_\beta$ and $i_\beta(X_\beta)$

Grade: A

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:

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