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Let $X_1\to Y$ and $X_2\to Y$ be two coverings. How does a morphism $X_1\to X_2$ over $Y$ induce morphism $\operatorname{Aut}_Y(X_1)\to \operatorname{Aut}_Y(X_2)$?

It should be trivial, but I can not understand it even in case of $Y$ point: if $X_1\to X_2$ is a mapping of sets, then how does it induce mapping $\operatorname{Aut}(X_1)\to \operatorname{Aut}(X_2)$?

evgeny
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  • $\operatorname{Aut}_Y(X_i)$ is the deck transformation group? – Daniel Fischer Jul 27 '14 at 13:08
  • @DanielFischer: Yes, it is. – evgeny Jul 27 '14 at 13:10
  • I think we need additional assumptions to get the induced map. Generally, I'd expect the induced mapping to be along the lines of "Fix $x_1\in X_1$, for $\varphi\in\operatorname{Aut}Y X_1$, let $f\ast(\varphi)$ be the deck transformation with $f_\ast(\varphi)(f(x_1)) = f(\varphi(x_1))$". For that to be well-defined, you need some conditions on the coverings (but it's been too long for me to remember which conditions). – Daniel Fischer Jul 27 '14 at 13:17

1 Answers1

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There is no such mapping in general. In your example where $Y$ is a point, you would like to extend $X \mapsto \operatorname{Sym}(X)$, the symmetric group, to a functor $S:\operatorname{Set} \to \operatorname{Group}$. Suppose there is such an $S$. Choose $X_1$ and $X_2$ with $|X_1| = 4$ and $|X_2| = 5$, and choose maps $f: X_1 \to X_2$ and $g: X_2 \to X_1$ with $g \circ f = 1$. The only nontrivial proper normal subgroup of $\operatorname{Sym}(X)$ is the alternating group when $|X| \ge 5$, so there are no onto homomorphisms $\operatorname{Sym}(X_2) \to \operatorname{Sym}(X_1)$. So $S(g)$ is not onto, and $S(g \circ f)$ is not onto. But $S(g \circ f) = S(1_{X_1}) = 1_{S(X_1)}$, so this is impossible. So there is no such $S$.

Hew Wolff
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