A map is continuous if and only if the pre-image of every closed set is closed

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Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be a mapping, then[1][2][3]:


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Continuity [ilmath]\implies[/ilmath] the pre-image of every closed set is closed

Let [ilmath]E[/ilmath] be closed in [ilmath](Y,\mathcal{ K })[/ilmath]

  • then [ilmath]f^{-1}(Y-E)\in\mathcal{J} [/ilmath] (as [ilmath]Y-E\in\mathcal{K} [/ilmath] and [ilmath]f[/ilmath] is continuous)
    • So [ilmath]X-f^{-1}(Y-E)[/ilmath] is closed in [ilmath]X[/ilmath]
      • But by properties of the pre-image of a function we have [ilmath]f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)[/ilmath] so:
        • [ilmath]X-f^{-1}(Y-E)=X-(f^{-1}(Y)-f^{-1}(E))[/ilmath], as [ilmath]f:X\rightarrow Y[/ilmath] we see [ilmath]f^{-1}(Y)=X[/ilmath], so:
          • [ilmath]=X-(X-f^{-1}(E))=f^{-1}(E)[/ilmath], thus we conclude:
            • [ilmath]f^{-1}(E)[/ilmath] is closed in [ilmath]X[/ilmath]
  • Since [ilmath]E[/ilmath] closed in [ilmath]Y[/ilmath] was arbitrary we have shown this is the case for all [ilmath]E[/ilmath]

the pre-image of every closed set is closed [ilmath]\implies[/ilmath] continuity

We wish to show [ilmath]f[/ilmath] is continuous, that is [ilmath]\forall U\in\mathcal{K}[f^{-1}(U)\in\mathcal{J}][/ilmath].

  • Let [ilmath]U\in\mathcal{K} [/ilmath] be given.
    • Then [ilmath]Y-U[/ilmath] is closed in [ilmath](Y,\mathcal{ K })[/ilmath]
      • By hypothesis [ilmath]f^{-1}(Y-U)[/ilmath] is closed in [ilmath](X,\mathcal{ J })[/ilmath]
  • Since [ilmath]U\in\mathcal{K} [/ilmath] was arbitrary we have shown the very definition of [ilmath]f[/ilmath] being continuous. As required.

TODO: Is "idempotent" the right for for describing [ilmath]X-(X-A)=A[/ilmath]?

See also


  1. Introduction to Topological Manifolds - John M. Lee
  2. Introduction to Topology - Bert Mendelson
  3. Topology - James R. Munkres