I posted this on MSE, but didn't get any responses, so I'm reposting here. I tried to write down an example of the main theorem of geometric class field theory, but I must be misunderstanding something, since my example isn't turning out right. **Main theorem.** Let $K$ be a function field of a curve over a finite field. There is a bijection between unramified $\ell$-adic Galois representations of $G_K := \operatorname{Gal}(K^{sep}/K)$ and $\ell$-adic characters of $K^{\times}\backslash\mathbf{A}_K^{\times}/\mathcal{O}_K^{\times}.$ (Here I mean the ideles modulo the diagonal on the left, and on the right quotiented by the subgroup of ideles which have valuation zero at every place.) -- I was trying to understand what this means when $K = \mathbb{F}_p(T).$ I think unramified Galois representations are the same as representations of $\operatorname{Gal}(K^{un}/K) = \operatorname{Gal}(\overline{\mathbf{F}_p}(T)/\mathbf{F}_p(T))= \hat{\mathbf{Z}}.$ So on one-side of the bijection, I have continuous $\ell$-adic characters of the profinite integers. -- On the other side, I think that this double quotient is just $\mathbb{Z}.$ One way is by identifying it with the Picard group of $\mathbb{P}^1$, but I also can do it more directly (thanks to these two 'independent' solutions, I don't think this is the incorrect step). -- So, I think that in this case, geometric class field theory is telling me that continuous homomorphisms $\rho : \hat{\mathbf{Z}} \to \overline{\mathbf{Q}_{\ell}}^{\times}$ are in bijection with continuous homomorphisms $\mathbf{Z}\to \overline{\mathbf{Q}_{\ell}}^{\times}.$ But this surely is incorrect -- the image of a map out of the profinite integers must be compact, but the image out of $\mathbf{Z}$ is not (take some $\ell$-adic number of multiplicative norm exceeding 1, for instance; this defines a map out of $\mathbb{Z}$ whose image is not contained in any compact set). So, where have I made a mistake?