Characteristic property of the disjoint union topology/Proof

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Suppose f:αIXαY is continuous βI[f|Xβ:XβY is continuous]

Let βI be given.

  • Let UK be given (so U is an open set in Y)
    • By hypothesis, f1(U)J (where J is the topology on αIXα)
    • This means γIVJγ[f1(U)iγ(Xγ)=V]
    • Thus VJβ[f1(U)iβ(Xβ)=V]
    • But:
      • f|1Xβ(U)=f1(U)Xβ Caution:This really ought to have its own proof
    • So for V being the image of some open set under the canonical injection, iβ

Work required:

  1. Some sort of homeomorphism between (Xα,Jα) and the subspace Xα:=iα(Xα) is needed