Characteristic property of the disjoint union topology/Proof

Suppose [ilmath]f:\coprod_{\alpha\in I}X_\alpha\rightarrow Y[/ilmath] is continuous [ilmath]\implies[/ilmath] [ilmath]\forall\beta\in I\big[f\big\vert_{X_\beta^*}:X_\beta^*\rightarrow Y[/ilmath] is continuous[ilmath]\big][/ilmath]

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

• Let [ilmath]U\in\mathcal{K} [/ilmath] be given (so [ilmath]U[/ilmath] is an open set in [ilmath]Y[/ilmath])
  • By hypothesis, [ilmath]f^{-1}(U)\in\mathcal{J}[/ilmath] (where [ilmath]\mathcal{J} [/ilmath] is the topology on [ilmath]\coprod_{\alpha\in I}X_\alpha[/ilmath])
  • This means [ilmath]\forall\gamma\in I\exists V\in\mathcal{J}_\gamma[f^{-1}(U)\cap i_\gamma(X_\gamma)=V][/ilmath]
  • Thus [ilmath]\exists V\in\mathcal{J}_\beta[f^{-1}(U)\cap i_\beta(X_\beta)=V][/ilmath]
  • But:
    • [ilmath]f\big\vert_{X_\beta^*}^{-1}(U)=f^{-1}(U)\cap X_\beta^*[/ilmath] Caution:This really ought to have its own proof
  • So for [ilmath]V[/ilmath] being the image of some open set under the canonical injection, [ilmath]i_\beta[/ilmath]

Work required:

1. Some sort of homeomorphism between [ilmath](X_\alpha,\mathcal{J}_\alpha)[/ilmath] and the subspace [ilmath]X_\alpha^*:=i_\alpha(X_\alpha)[/ilmath] is needed