Disjoint union topology

From Maths
(Redirected from Disjoint union (topology))
Jump to: navigation, search
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:
Grade A until it is more presentable.

Definition

Suppose [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be an indexed family of topological spaces that are non-empty[1], the disjoint union topology is a topological space:

  • with underlying set [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath], this is the disjoint union of sets, recall [ilmath](x,\beta)\in\coprod_{\alpha\in I}X_\alpha\iff \beta\in I\wedge x\in X_\beta[/ilmath] and
  • The topology where [ilmath]U\in\mathcal{P}(\coprod_{\alpha\in I}X_\alpha)[/ilmath] is considered open if and only if [ilmath]\forall \alpha\in I[X_\alpha\cap U\in\mathcal{J}_\alpha][/ilmath][Note 1] - be sure to notice the abuse of notation going on here.

TODO: Flesh out notes, mention subspace [ilmath]X_\alpha\times\{\alpha\} [/ilmath] and such


Claim 1: this is indeed a topology


TODO: Define the canonical injections of the disjoint union topology here


Note that the canonical injections of the disjoint union topology are topological embeddings

Characteristic property

Let [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] be a collection of topological spaces and let [ilmath](Y,\mathcal{ K })[/ilmath] be another topological space]]. We denote by [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath] the disjoint unions of the underlying sets of the members of the family, and by [ilmath]\mathcal{J} [/ilmath] the disjoint union on it (so [ilmath](\coprod_{\alpha\in I}X_\alpha,\mathcal{J})[/ilmath] is the disjoint union topological construct of the [ilmath]\big((X_\alpha,\mathcal{J}_\alpha)\big)_{\alpha\in I} [/ilmath] family) and lastly, let [ilmath]f:\coprod_{\alpha\in I}X_\alpha\rightarrow Y[/ilmath] be a map (not necessarily continuous) then:
we state the characteristic property of disjoint union topology as follows:


TODO: rewrite and rephrase this


  • [ilmath]f:\coprod_{\alpha\in I}X_\alpha\rightarrow Y[/ilmath] is continuous if and only if [ilmath]\forall\alpha\in I\big[f\big\vert_{X_\alpha^*}:i_\alpha({X_\alpha})\rightarrow Y\text{ is continuous}\big][/ilmath]

Where (for [ilmath]\beta\in I[/ilmath]) we have [ilmath]i_\beta:X_\beta\rightarrow\coprod_{\alpha\in I}X_\alpha[/ilmath] given by [ilmath]i_\beta:x\mapsto(\beta,x)[/ilmath] are the canonical injections

Proof of claims

(Unknown grade)
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:
Actually surprisingly easy to prove, done on paper. page 1, 7/8/2016, Intro to top manifolds notes. Filed

Notes

  1. There's a very nasty abuse of notation going on here. First, note a set [ilmath]U[/ilmath] is going to be a bunch of points of the form [ilmath](x,\gamma)[/ilmath] for various [ilmath]x[/ilmath]s and [ilmath]\gamma[/ilmath]s ([ilmath]\in I[/ilmath]). There is no "canonical projection" FROM the product to the spaces, as this would not be a function!

References

  1. Introduction to Topological Manifolds - John M. Lee



TODO: Investigate the need to be non-empty, I suspect it's because the union "collapses" in this case, and the space wouldn't be a part of union