Are commutative C*-algebras really dual to locally compact Hausdorff spaces?

It is true that the category of locally compact Hausdorff spaces is equivalent to the category of commutative $C^*$-algebras . . . with appropriately chosen morphisms.

Let $A$ and $B$ be commutative $C^*$-algebras. Then, a morphism from $A$ to $B$ is defined to be a nondegenerate homomorphism of $^*$-algebras from $\phi :A\rightarrow M(B)$, where $M(B)$ is the multiplier algebra of $B$. Here, nondegeneracy means that the the span of $\left\{ \pi (a)b:a\in A,b\in M(B)\right\}$ is dense in $M(B)$. Note that you need a bit of machinery to even make this into a category because it is not obvious a priori that composition makes sense. Nevertheless, it does work out. Proposition 1 on pg. 11 and Theorem 2 on pg. 12 of Superstrings, Geometry, Topology, and $C^*$-algebras (in fact this chapter is on the arXiv) respectively show that this forms a category and that the dual of this category is equivalent to the category of locally compact Hausdorff spaces.

So, for the sake of having an answer written down to this question: as t.b. says in the comments, we simply don't have this duality as stated.

The following categories are contravariantly equivalent:

• locally compact Hausdorff spaces with proper continuous maps
• commutative C$^*$-algebras with non-degenerate $*$-homomorphisms

Here, a $*$-homomorphism $f : A \to B$ is non-degenerate if the following equivalent conditions are satisfied:

1. The ideal generated by the set-theoretic image of $f$ is dense in $B$.
2. For every approximative unit $(u_i)$ in $A$ its image $f(u_i)$ is an approximative unit in $B$.
3. For some approximative unit $(u_i)$ in $A$ its image $f(u_i)$ is an approximative unit in $B$.

I don't think that multiplier algebras are necessary ...