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I have seen it claimed, for example here (in the case $n=2$) that if $k$ is a field and $f_1,\ldots,f_n \in k[[x_1,\ldots,x_n]]$ all of which have zero constant coefficient, the homomorphism

$$\phi: k[[x_1,\ldots,x_n]] \to k[[x_1,\ldots,x_n]] $$ $$ \phi(x_i) = f_i$$

is an isomorphism if and only if $$\det\left(\frac{\partial f_i}{\partial x_j}\right)_{i,j} $$ is invertible i.e. has non-zero constant coefficient.

I am primarily interested in a reference where I can find a proof of this result.

This question has been asked here, but it is marked as a duplicate and the link to the question which it is a duplicate of is dead.

Partial progress: Surjectivity is not too difficult. Let $$J(x_1,\ldots,x_n) = \left(\frac{\partial f_i}{\partial x_j}\right) $$ and let $H(x_1,\ldots,x_n)$ be the inverse matrix. Then $$\sum_{j=1}^n H_{ij}(x_1,\ldots,x_n)f_i(x_1,\ldots,x_n) = x_i + \text{ higher order terms }$$ Then we can make successive higher-order corrections to get an expression for $x_i$. Hence, $x_1,\ldots,x_n$ are in the image of the map $\phi$.

ducksforever
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    The duplicate has been updated - see here for the original answer which contains a full solution. (I thought the site would not allow deletion of an intermediate entry in a chain of duplicates, but apparently it can happen sometimes.) – KReiser Apr 18 '23 at 20:37
  • @KReiser Thank you for linking to the original question. Would it be appropriate to delete my question, given that it isn't adding much to the site? – ducksforever Apr 18 '23 at 20:44
  • If your question is answered fully, then sure, feel free to delete. If there's something that's still unanswered, I would recommend editing your post to make that clear. – KReiser Apr 18 '23 at 20:48

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