# Epimorphisms of locally compact spaces

Let $LCH$ be the category of locally compact Hausdorff spaces with proper continuous maps.

Question. What are the epimorphisms in $LCH$?

I suspect them to be surjective, but I haven't been able to prove it.

Here is an idea: Let $f : X \to Y$ be an epimorphism. The image $f(X)$ is closed. It follows that $f(X)^+$ is closed in $Y^+$, where $+$ denotes the Alexandrov compactification. If $y \in Y \setminus f(X)$, by Urysohn's Lemma there is some $g \in C(Y^+)$ with $g(y)=1$ and $g(f(X)^+)=0$. Now, $g(\infty)=0$ implies that $g$ restricts to some $g \in C_0(Y)$ such that $g(y)=1$ and $g(f(X))=0$. The remaining problem is that $g$ might be not proper.

I have also tried to prove it in the dual category, which is the category $CommC^*Alg$ of commutative $C^*$-algebras with non-degenerate $*$-homomorphisms. I already know that surjections become injections under this duality, so that the question would be: Is every monomorphism in $CommC^*Alg$ injective? Again, the restrictive morphisms cause some problems in proving this.

• They are epic! :-) – Asaf Karagila May 16 '15 at 9:47

Yes, a morphism in this category is epic iff it is surjective. Obviously surjective morphisms are epic, so it suffices to show that if $f: X\to Y$ is not surjective, there is a locally compact Hausdorff space $Z$ and distinct $g_1, g_2 \in \hom(Y,Z)$ such that $g_1 \circ f = g_2 \circ f$.
On the product $Y \times 2$ define the equivalence relation $E$ such that $(x, i) \,E\, (y, j)$ iff $x = y$ and ($i = j$ or $x \in f(X)$). (In other words: glue the two copies of $f(X)$ together). Let $Z = (Y \times 2)/E$ and $q: Y \times 2 \to Z$ the natural map. Since the fibres of $q$ are finite they are compact and, using the fact that $f(X)$ is closed, it is easy to verify that if $F$ is closed in $Y \times 2$, so is $q^{-1}(q(F))$. Hence $q$ is perfect and since perfect images of locally compact Hausdorff spaces are also locally compact Hausdorff (thm. 3.7.20, 3.7.21), $q$ is a morphism.
If we now take $g_1$ and $g_2$ to be the compositions of $q$ with the two natural embeddings $Y \to Y \times 2$, we find they differ on $Y\setminus f(X)$, but agree on $f(X)$, therefore $g_1 \circ f = g_2 \circ f$.