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

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## Contents

## 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]}:

- [ilmath]f:X\rightarrow Y[/ilmath] is continuous
*if and only if*[ilmath]\forall E\in C(\mathcal{K})[f^{-1}(E)\in C(\mathcal{J})][/ilmath] (where [ilmath]C(\mathcal{H})[/ilmath] denotes the*set of all*closed sets for a topology [ilmath]\mathcal{H} [/ilmath])

## 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]

- [ilmath]=X-(X-f^{-1}(E))=f^{-1}(E)[/ilmath], thus we conclude:

- [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:

- 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:

- So [ilmath]X-f^{-1}(Y-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]
- So [ilmath]X-f^{-1}(Y-U)\in\mathcal{J} [/ilmath] (is open in [ilmath](X,\mathcal{ J })[/ilmath])
- But [ilmath]X-f^{-1}(Y-U)=X-(X-f^{-1}(U))=f^{-1}(U)[/ilmath] by properties of the pre-image of a function and relative complement is idempotent
- So [ilmath]f^{-1}(U)[/ilmath] is open in [ilmath](X,\mathcal{ J })[/ilmath]

- But [ilmath]X-f^{-1}(Y-U)=X-(X-f^{-1}(U))=f^{-1}(U)[/ilmath] by properties of the pre-image of a function and relative complement is idempotent

- So [ilmath]X-f^{-1}(Y-U)\in\mathcal{J} [/ilmath] (is open in [ilmath](X,\mathcal{ J })[/ilmath])

- By hypothesis [ilmath]f^{-1}(Y-U)[/ilmath] is closed in [ilmath](X,\mathcal{ J })[/ilmath]

- Then [ilmath]Y-U[/ilmath] is closed in [ilmath](Y,\mathcal{ K })[/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

## References

- ↑ Introduction to Topological Manifolds - John M. Lee
- ↑ Introduction to Topology - Bert Mendelson
- ↑ Topology - James R. Munkres