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It's clear that the étale topology is closed related to some form of inverse function theorem. Let me give some reasons. (Also, see the comments here.)

(1) A morphism $f:X\to S$ between smooth varieties over $\bar{k}$ is étale at $p\in X$ if and only if $\mathrm{d}f_p:T_pX\to T_{f(p)}S$ is an isomorphism.

(2) A smooth morphism $f:X\to S$ of relative dimension $n$ is, étale locally, an open immersion into $\mathbb{A}_S^n$. (This is basically the structure theorem of submersions.)

(3) If $f:X\to S$ is finite étale, then every $s\in S$ has an étale neighborhood $V\to S$ such that $X\times_S V\to V$ is a trivial cover.

(4) If $\bar{x}$ is a geometric point of a scheme $X$, the ring $\mathcal{O}_{X,\bar{x}}$ is strictly henselian. This is the ring of germs of functions étale locally at $\bar{x}$, and the Hensel lemma is very closed related to the inverse function theorem.

However, I still don't know what's precisely the inverse function theorem in this context. I would love if it were the following statement, which is clearly not true.

Let $f:X\to S$ be a morphism between smooth varieties over $\bar{k}$ such that $\mathrm{d}f_p$ is an isomorphism. Then there exists an étale neighborhood $V\to S$ of $f(p)$ such that $X\times_S V\to V$ is an isomorphism.

P.S.: I know that this question is perhaps too simple for MO, and that it's similar to this. I still think that here's the best place and that it's sufficiently different to the other question. Feel free to close the question, if you think otherwise.

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    $\begingroup$ The last statement you ask about is false even in the differentiable category (just take any non-trivial covering space). If you allow the source to be replaced by an étale neighborhood of $p$ as well, this makes the statement true and a trivial consequence of your property (1). $\endgroup$ Commented Jan 19, 2022 at 17:32
  • $\begingroup$ @DenisNardin Could you give more details about why this is trivial if we replace the source as well? I get that (1) is used for proving that $f$ is étale, but I don't see how to find the étale neighborhood of $p$. $\endgroup$
    – Gabriel
    Commented Jan 19, 2022 at 17:40
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    $\begingroup$ This is meant to be a tautology. If $df_p$ is an isomorphism, i.e. $f$ is etale at $p$, then there exists an open neighborhood $V$ of $p$ such that $f|_V$ is etale. This $V$ is then an etale neighborhood of $f(p)$, and so $f$ tautologically gives an isomorphism from $V$ (a neighborhood of $p$) to $V$ (a neighborhood of $f(p)$). $\endgroup$ Commented Jan 19, 2022 at 17:46
  • $\begingroup$ Have you had a look at math.stackexchange.com/questions/3314133/… ? $\endgroup$ Commented Feb 4, 2022 at 8:45

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