View source for Characteristic property of the disjoint union topology/Proof

==Suppose {{M|f:\coprod_{\alpha\in I}X_\alpha\rightarrow Y}} is continuous {{M|\implies}} {{M|1=\forall\beta\in I\big[f\big\vert_{X_\beta^*}:X_\beta^*\rightarrow Y}} is continuous{{M|1=\big]}}==
Let {{M|\beta\in I}} be given.
* Let {{M|U\in\mathcal{K} }} be given (so {{M|U}} is an [[open set]] in {{M|Y}})
** By hypothesis, {{M|1=f^{-1}(U)\in\mathcal{J} }} (where {{M|\mathcal{J} }} is the [[topology]] on {{M|\coprod_{\alpha\in I}X_\alpha}})
** This means {{M|1=\forall\gamma\in I\exists V\in\mathcal{J}_\gamma[f^{-1}(U)\cap i_\gamma(X_\gamma)=V]}}
** Thus {{M|1=\exists V\in\mathcal{J}_\beta[f^{-1}(U)\cap i_\beta(X_\beta)=V]}}
** But:
*** {{M|1=f\big\vert_{X_\beta^*}^{-1}(U)=f^{-1}(U)\cap X_\beta^*}} {{Caution|This really ought to have its own proof}}
** So {{M|f\big\vert_{X_\beta^*}^{-1}(U)=V}} for {{M|V}} being the image of some open set under the canonical injection, {{M|i_\beta}}
Work required:
# Some sort of [[homeomorphism]] between {{M|(X_\alpha,\mathcal{J}_\alpha)}} and the {{link|subspace|topology}} {{M|1=X_\alpha^*:=i_\alpha(X_\alpha)}} is needed