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Think of the contravariant functor sending a non-unital commutative Banach algebra to its character space

$$F : \mathcal A^{\text{op}} \to \operatorname{LCHaus}, A \mapsto \hat A; f^{\text{op}} : A \to B \mapsto \cdot \circ f : \hat A \to \hat B $$

What are the "right" morphisms in $\mathcal A$? Characters need to be non-zero. This disqualifies some of the the typical, say, short (i.e. norm-decreasing) algebra morphisms. To ensure that $\tau\circ f$ is always a character we may want to insist that $f$ is surjective. But I'm not sure that's the best choice we have.

Stefan Perko
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  • If you write the opposite, it becomes a covariant functor. – J. De Ro Sep 08 '22 at 08:09
  • Regardless of the morphisms in $\mathcal{A}$, the space of characters can be zero. – J. De Ro Sep 08 '22 at 08:11
  • @QuantumSpace A matter of notation I suppose... "Can be zero" - do you mean empty? It would indeed be problem if $\hat B$ is empty, but $\hat A$ isn't. – Stefan Perko Sep 08 '22 at 08:14
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    I asked a related question a long time ago here which got a lot of great answers: https://math.stackexchange.com/questions/170984/are-commutative-c-algebras-really-dual-to-locally-compact-hausdorff-spaces – Qiaochu Yuan Sep 08 '22 at 08:17
  • @QiaochuYuan Oh boy. That's really what I was looking even if I didn't know at all what to expect. – Stefan Perko Sep 08 '22 at 08:21
  • @StefanPerko Yes I meant empty, sorry (if you require characters to be non-zero, otherwise there is the zero character). Also the related question is somewhat different: if $A$ is a commutative $C^*$-algebra, then $A=C_0(X)$ for some locally compact Hausdorff space and you have lots of characters. For a general commutative Banach algebra, nothing ensures one character even exists. – J. De Ro Sep 08 '22 at 08:25
  • @QuantumSpace Ah true, I was not reading it very carefully. – Stefan Perko Sep 08 '22 at 08:27

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