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A continuous map of compact Hausdorff spaces is an ultrafilter-algebra homomorphism

Statement

Let f:XY be continuous between compact Hausdorff spaces, and let ξX,ξY be their ultrafilter-limit maps. Then

fξX=ξYβ(f),

so f is an ultrafilter-algebra homomorphism.

Facts & Assumptions

Given: A continuous map f:XY of compact Hausdorff spaces and an ultrafilter U on X.

[L1]

The pushforward fU is an ultrafilter on Y, and ultrafilter pushforward is functorial (Pushforward sends ultrafilters to ultrafilters and is functorial).

[L2]

Every ultrafilter on a compact Hausdorff space has exactly one limit (A given ultrafilter on a compact Hausdorff space has a unique limit).

[L3]

A T-algebra homomorphism f:(A,a)(B,b) satisfies fa=bT(f) (Algebra and algebra homomorphism for a monad).

Proof

technique · direct
1.1

Push U forward along f to the ultrafilter fU on Y supplied by [L1].

L1
2.1

If U converges to x, then for each neighbourhood V of f(x), continuity makes f1[V] a neighbourhood of x and hence a member of U. Therefore VfU, so the pushforward converges to f(x).

step 1.1given
3.1

Taking x=ξX(U), uniqueness in the target gives ξY(fU)=f(ξX(U)).

step 2.1L2
4.1

Since fU=β(f)(U), step 3.1 is the equation fξX=ξYβ(f) in [L3]. Thus f is an algebra homomorphism, including for the unique map from an empty compact space.

step 3.1L3

Depends on

Used by

Dependency tree · two levels

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