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My book is An Introduction to Manifolds by Loring W. Tu.

The last sentence here says

being an immersion or a submersion at p is equivalent to the maximality of rk $[\frac{\partial f^i}{\partial x_j}(p)]$

Is the above sentence then a generalization of the inverse function theorem, which is Theorem 6.26 and Remark 8.12 and which I asked about here, when we allow $\dim M \ne \dim N$? I think "full rank" is a generalization of "invertible" when our dimensions are not equal. Therefore, the sentence for equal dimensions gives "immersion or submersion" as "immersion at $p$, or equivalently submersion at $p$, or equivalently $F_{*,p}$ is an isomorphism" and "full rank" as "$F_{*,p}$, the map, or equivalently $F_{*,p}$, the Jacobian matrix, is invertible".

Edit: I edited the last paragraph and the title by asking about generalizations instead of analogies. I think the sentence is merely an analogy if we're given $\dim M \ne \dim N$ but a generalization if we allow both $\dim M \ne \dim N$ and $\dim M = \dim N$.

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    You are looking for the implicit function theorem which is very closely related to the inverse function theorem. – Lutz Lehmann Jul 25 '19 at 05:27
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    ^ yes you're probably looking for the implicit function theorem (funny how both are IFT). In fact they're equivalent in the sense that inverse function theorem implies implicit function theorem and vice-versa. – peek-a-boo Jul 25 '19 at 05:30
  • @LutzL Thanks. Edited to $\dim N$. As for implicit function theorem, the one in the book I think assumes equal dimensions. Later the book says "The implicit function theorem is equivalent to the inverse function theorem" rather than "The implicit function theorem is equivalent to the inverse function theorem, for $n=m$." Is your version of implicit function theorem different? –  Jul 25 '19 at 05:32
  • @peek-a-boo Thanks. The implicit function theorem in the book I think assumes equal dimensions. Later the book says "The implicit function theorem is equivalent to the inverse function theorem" rather than "The implicit function theorem is equivalent to the inverse function theorem, for $n=m$." Is your version of implicit function theorem different? –  Jul 25 '19 at 05:32
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    If $F:\Bbb R^n\to\Bbb R^m$, $m<n$ has a full-rank Jacobian as some point $x_0$, then you can select a matrix $A$ consisting of basis vectors so that the Jacobian of $(F,A):\Bbb R^n\to\Bbb R^n$ is invertible at $x_0$ and you can apply the inverse function theorem solving $F(x)=y,~Ax=z$ for $x\approx x_0$, $y\approx F(x_0)$, $z\approx Ax_0$. As $z$ is a selection of coordinates of $x$, this prescribes how to find the other coordinates in dependence of them, which is what the implicit function theorem says. – Lutz Lehmann Jul 25 '19 at 05:40
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    no I'm referring to the same theorem. clearly in the implicit function theorem the dimensions are different. the domain is an open subset of $m+n$ dimensional space while the target space is only $m$ dimensional. also, there is no inverse function theorem for different dimensions in the domain and target space. What I mean is that if $f:V \to W$ is a differentiable mapping between normed vector spaces with differentiable inverse, then $V$ and $W$ are isomorphic (apply the chain rule to $f\circ f^{-1} = \text{id}_W$) – peek-a-boo Jul 25 '19 at 05:40
  • @peek-a-boo But it says "determinant" in it. Determinant is for same dimensions. What's going on? Also is the sentence then equivalent to implicit function theorem? –  Jul 25 '19 at 05:41
  • @LutzL Ok I'll take your word for it. Is the sentence then equivalent to implicit function theorem? –  Jul 25 '19 at 05:42
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    I'm not sure about "equivalent", but in both cases it is a direct consequence of the implicit function theorem. Both theorems essentially say that in the case of full rank of the Jacobian, the non-linear system is locally solvable in the same way as the linear system with the Jacobian as matrix, the same sub-divisions into free and dependent variables apply, the non-linear solution can be obtained as small perturbation of the linear solution. – Lutz Lehmann Jul 25 '19 at 05:45
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    Perhaps you're misunderstanding the word equivalent. What we mean is that if you assume the implicit function theorem is true, then you can prove the Inverse function theorem (and vice versa). also, in the implicit function theorem, it talks about the determinant of a specific $m \times m$ submatrix, soo I'm not sure yet exactly your confusion is about. before trying to seek any form of generalisation to these theorems, i suggest you carefully read their statements and understand them, and if needed, rephrase/ask another question about any doubts you have (be specific about it) – peek-a-boo Jul 25 '19 at 05:49
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    @peek-a-boo : I was answering to your last statement which implied an equivalence of the IFT (whatever version) and the immersion/submersion claim. I can see that they are closely enough related, in the way that the extended mean value theorem is equivalent to the Rolle theorem, I'm just not sure about full equivalence, and if the difference in geometric content makes them too different. – Lutz Lehmann Jul 26 '19 at 14:04
  • @LutzL The analogy is that the extended mean value theorem (I guess you mean Cauchy) is the sentence (the "immersion/submersion claim") and Rolle's theorem is the inverse function theorem? –  Jul 31 '19 at 06:36
  • @peek-a-boo I didn't study implicit function theorem, but I studied inverse function theorem. The latter IFT is what my question is (directly) about...Anyway, what's the relation between the sentence given and either of the IFT's please? Based on Remark 8.12, I really do believe the sentence is an unequal dimension version for IFT...in particular a generalization of IFT which allows for equal or unequal. –  Jul 31 '19 at 06:37
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    I'm afraid I still do not see your concerns clearly. The inverse function theorem only holds when both dimensions are the same, because otherwise the push-forward map $F_{*, p} $ cannot be an isomorphism of vector spaces (unequal manifold dimensions implies unequal tangent space dimensions). Also, Intuitively, it should make sense that (after assuming enough "niceness" ) you can only have an inverse which maps between spaces of same dimension. so there is no inverse function theorem when the spaces havr different dimensions. – peek-a-boo Jul 31 '19 at 15:01
  • @peek-a-boo Thanks. Never mind. I edited the question from analogue to generalization. Oh but actually your comment is still the same I think? –  Aug 01 '19 at 03:04
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    yup, my comment still holds: you need to have the same dimension on both the domain and target space. – peek-a-boo Aug 01 '19 at 03:09
  • @peek-a-boo Thanks. When I say inverse function theorem, I refer to Remark 8.12. I believe Remark 8.12 is equivalent to "immersion and submersion if and only if full rank and same dimension". I now generalize this "immersion or submersion if and only if full rank". Is any part of the preceding wrong? –  Aug 01 '19 at 03:32

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