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Let $k$ be an algebraically closed field and $f\colon X \rightarrow Y$ an etale morphism of smooth curves over $k$. Let $f^{[n]}\colon X^{[n]} \rightarrow Y^{[n]}$ be the induced morphism on $n$-th symmetric powers. Is $f^{[n]}$ also etale? Clearly, affine $X$ and $Y$ suffice for the purposes of the question.

It seems from VA.'s answer to Smoothness of Symmetric Powers that the answer is yes, but his argument is too terse for me; in particular, I don't understand how to harmlessly pass to formal completions. I would be grateful if someone could give a more detailed explanation.

EDIT. As abx noted in the comments, my question is based on a false expectation. Give this, let me modify the question: how does one prove that $X^{[n]}$ is $k$-smooth by reducing to the $X = \mathbb{A}^1_k$ case? That is, how to carry out the reduction?

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    $\begingroup$ This is obviously false. Say $f$ has degree 2, and $f^{-1}(y)=\{x_1,x_2\} $; then the fiber of $f^{[2]}$ above $(y,y)$ consists of 3 points $(x_1,x_1)$, $(x_2,x_2)$ and $(x_1,x_2)$, while $f^{[2]}$ has degree 4. $\endgroup$
    – abx
    Commented Jan 19, 2015 at 7:14
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    $\begingroup$ As abx said, this is false. A slightly expanded explanation is that it is true for product, but not symmetric product. That's because the symmetric group doesn't act freely. $\endgroup$
    – Charles Siegel
    Commented Jan 19, 2015 at 7:46
  • $\begingroup$ Answer to the EDIT question: locally for the étale topology, $X$ is isomorphic to $\mathbb{A}^1$, hence $X^{[n]}$ is isomorphic to $(\mathbb{A}^1)^{[n]}\cong \mathbb{A}^n$, and therefore smooth. $\endgroup$
    – abx
    Commented Jan 19, 2015 at 16:52
  • $\begingroup$ @abx: The "hence" part of your comment is precisely what I am asking to explain. I understand the general strategy that you mention, but my question is: how to make it precise? $\endgroup$
    – Lisa S.
    Commented Jan 19, 2015 at 16:57
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    $\begingroup$ @abx: I have already tried to do this along the lines you indicate but I don't find it easy. That is why I am asking this here. Your comments do not answer my question. $\endgroup$
    – Lisa S.
    Commented Jan 19, 2015 at 17:18

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