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Let $X=\mathbf{A}^1_{\overline{\mathbf{Q}}}-\{0\} = \mathbf{G}_{m,\overline{\mathbf{Q}}}$ be the multiplicative over the field of algebraic numbers. Each finite etale cover $Y\to X$ (with $Y$ connected) is isomorphic to the finite etale morphism $X\to X$ given by $z\mapsto z^n$ for some $n\geq 1$.

The universal covering space $\widetilde{X}$ of $X$ is the projective limit over all finite etale covers of $X$. It's not a scheme. (If $\widetilde{X}$ were a scheme, the "morphism" $\widetilde{X}\to X$ would be an etale morphism with non-finite fibres. That's not possible.)

Is it endowed with a morphism $\widetilde{X}\to X$? In which category should I consider this "morphism"?

Can we describe $\widetilde{X}$ a bit more explicitly using the above description of all finite etale covers of $X$?

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Since you have taken a projective limit, the map is probably in the pro-category which is just a formal construction where essentially the objects are projective systems and morphisms are maps that make all diagrams commute. –  Matt May 8 '12 at 20:07
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Alternatively, one could pass to an appropriate completion of such a category: let $\mathcal{C}$ be the category of schemes of finite presentation over $X$, and consider the category of presheaves $\widehat{\mathcal{C}} = [\mathcal{C}^\textrm{op}, \textbf{Set}]$; the Yoneda lemma implies that the Yoneda embedding $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}}$ is full and faithful, so you have $X$ sitting in this larger category $\widehat{\mathcal{C}}$; but $\widehat{\mathcal{C}}$ contains all small limits, so you can form the presheaf which would "represent" $\widetilde{X}$ too. –  Zhen Lin May 8 '12 at 21:48

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up vote 6 down vote accepted

One can form the projective limit $\tilde{X}$ in the category of schemes without any problem; it is Spec $\overline{\mathbb Q}(\{z^{1/n}\}_{n \geq 1})$. It is not finite type over $X$, and so in particular is not etale, but so what? There is no theorem saying that a projective limit of finite etale maps is etale.

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