Gelfand-Kolmogorov Theorem for the space $C(X)$ with compact $X$: ring vs algebra version.
I recalled a reference from a classic:
Exercise 1I in Gillman and Jerison's book Rings of continuous functions says:
let $\mathfrak{t}$ be a (ring) homomorphism from $C(Y)$ or $C^\ast(Y)$ (bounded continuous real-valued functions) into $C(X)$.
- $\mathfrak{t}\mathbf{r} = \mathbf{r}\cdot \mathfrak{t}\mathbf{1}$ for each $ r \in \Bbb R$ (where $\mathbf{r}$ is the constant function with value $r$). With hint
For each $x \in X$, the mapping $x \to (\mathfrak{t}\mathbf{r})(x)$ is a homomorphism from $\Bbb R$ into $\Bbb R$, and hence is either the zero homomorphism or the identity (0.22). So $(\mathfrak{t}\mathbf{1})(x) = 0$ or $1$.
- $\mathfrak{t}$ is an algebra homomorphism, i.e. $\mathfrak{t}(\mathbf{r}g) = \mathbf{r}\cdot \mathfrak{t}(g)$ for all $ r \in \Bbb R$ and $g \in C(Y)$.
I'll refrain from commenting on the categorial part.