Disjoint union topology

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Grade A until it is more presentable.

Definition

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

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

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TODO: Flesh out notes, mention subspace $X_\alpha\times\{\alpha\}$ and such

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Claim 1: this is indeed a topology

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

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

Characteristic property

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

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TODO: rewrite and rephrase this

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

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

Proof of claims

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

References

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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

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