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

## Statement

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]:

## Proof

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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]?