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Let $F: N \rightarrow M $ be a smooth map from smooth manifold $N$ of dimension n to smooth manifold $M$ of dimension m. $F$ is an immersion at $p \in N $ if $F_{*,p}$ is injective.

However $F_{*,p}$ dependents of the charts at $p$: $(U, \phi = (x^1,...,x^n))$ and at $\phi(p)$: $(V, \psi = (y^1,...,y^m))$. Concretely we have to calculate the rank of Jacobi matrix:

$J = (\frac{\partial F^i}{\partial x^j}) = (\frac{\partial r^i\circ\psi\circ F\circ\phi^{-1}}{\partial r^j})$ at $\phi(p)$

For immersion, it requires $rank(J)=n$

Whether could different choices of $\phi$ and $\psi$ lead to different values of $rank(J)$?

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Different choices of $\phi$ and $\psi$ would just lead to multiplying $J$ on the left or right by an invertible Jacobian. This is because of the chain rule and the fact that you're just doing a change of basis.

Consider the map $\psi\circ F\circ \phi^{-1}$. If we have $\phi'$ and $\psi'$ maps for different charts (with appropriate domains), we have that $$ \psi'\circ F\circ \phi'^{-1}=(\psi'\circ\psi^{-1})\circ(\psi\circ F\circ \phi^{-1})\circ(\phi\circ\phi'^{-1}). $$ Since $\phi$ and $\phi'$ are compatible (similarly, $\psi$ and $\psi'$ are compatible), the Jacobians for $\psi'\circ\psi^{-1}$ and $\phi\circ\phi'^{-1}$ are square and invertible. Hence, the injectivity (or noninjectivity) is preserved.

More precisely, by the chain rule $$ (\psi'\circ F\circ \phi'^{-1})_\ast=(\psi'\circ\psi^{-1})_\ast\circ(\psi\circ F\circ \phi^{-1})_\ast\circ(\phi\circ\phi'^{-1})_\ast. $$

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  • $\begingroup$ Thanks @MichaelBurr. It's clear to me that if two charts are compatible, then they will not change the immersion property of a map. However how about the two charts not being compatible? For example, if N is R (real axis) with single chart (R, id: id(x)=x), and M is R^2, then the identity map i: i(x) = (x,0) from N to M is clearly an immersion. However if N is R with single chart (R, x^(1/3)), then i is not a injective at x = 0. Whether does it mean the immersion of i depending on the chart selection? (The two charts are not compatible with each other) $\endgroup$
    – jizhou
    May 3 '16 at 14:37
  • $\begingroup$ Two charts which are not compatible come from different atlases. Since a manifold is a set $X$ and a choice of maximal atlas $M$, by using different atlases, you're changing the manifold. When you change the manifold, of course the immersions change because the differential structure changes. $\endgroup$ May 3 '16 at 14:44
  • $\begingroup$ Because a dimension 1 topological manifold only has one smooth structure (up to diffeomorphism), the two manifolds are diffeomorphic to each other. Are they different in this sense? And in the answer by @hase_olaf, the immersion property of a map doesn't depend on the chart selection. Is this a conflict? $\endgroup$
    – jizhou
    May 3 '16 at 15:01
  • $\begingroup$ You should probably write these as new questions. For the one smooth structure, part, you need to precompose with the diffeomorphism to preserve the immersion. After diffeomorphism, the map changes. In other words, if $F$ is the diffeomorphism and $i$ is the immersion, you need $F^\ast i$. $\endgroup$ May 3 '16 at 15:14
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    $\begingroup$ Thanks very much! I think I have realized the key point: manifolds with different smooth structures are different smooth manifolds. I also have understood the solution of @hase_olaf is actually equivalent to yours. To see the equivalence has frustrated me for whole night! $\endgroup$
    – jizhou
    May 4 '16 at 13:39
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I just want to add something to the calculation already performed by others:

$F$ is a map between the manifolds $N$ and $M$ and it doesn't care about charts. View the two manifolds just as sets, so $F$ assigns to every element of $N$ an element of $M$, independently of any chart. Similaraly $F_{*,p}$ is a map between the tangential planes $T_p N$ and $T_{F(p)} M$, that is defined without charts. [For example represent a given tagential vector $X \in T_p N$ by a curve $\gamma: [-a,a] \to N$ (meaning $\gamma (0) = p$ and $\dot{\gamma} (0) = X$), than $F_{*,p} X = \frac{d}{dt} |_{t=0} (F \circ \gamma)$.] So for $F$ to be an immersion at $p$, you need to make sure, that two different vectors $Y', Y'' \in T_p N$, $Y' \neq Y''$ go to different vectors in $T_{F(p)} M$, in other words: $F_{*,p} Y' \neq F_{*,p} Y''$. Formulated this way, the statement of the property doesn't use charts and is thus independent of charts.

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  • $\begingroup$ Thanks @hase_olaf for providing another view to this question. I just want to comment that when saying "to find a vector $Y' \in T_pN$" we have implicitly used chart. The same tangent vector in different charts have different forms. However since different smooth structures mean different smooth manifolds, no change is needed to your answer. $\endgroup$
    – jizhou
    May 4 '16 at 13:49
  • $\begingroup$ Sorry, I was so obsessed with submersions from my own stuff, that I plugged in the property of a submersion instead of an immersion. I'll edit the post. $\endgroup$
    – hase_olaf
    May 4 '16 at 16:08
  • $\begingroup$ The tangential plane at $p \in N$ is a vector space. Different charts give different bases for this vector space, with respect to which the components of a given tangential vector differ. Is this what you mean by "different forms"? But the vector stays the same. Same thing for $T_{F(p)} M$. Now $F_{*,p}$ is a linear map between theese two vector spaces. Its representation as a matrix depends on the choice of charts, but the map itself doesn't. So no, there is no implicit usage of charts. Do you know ways to look at tangent vector without using charts and the bases coming with them? $\endgroup$
    – hase_olaf
    May 4 '16 at 17:52
  • $\begingroup$ For example, let $N$ be $R$ equipped with single chart $(R, id: id(x)=x)$, and $M = R^2$, obviously the inclusion map $i: N \rightarrow M, i(x)=(x,0)$ is an immersion. If $N'$ is $R$ equipped with single chart $(R, \phi (x)=x^{\frac{1}{3}})$, then the inclusion map $i': N' \rightarrow M, i'(x)=(x,0)$ is not an immersion. If we don't consider chart, then $N$ is the same as $N'$. However at $x=0$ in $N'$, the only tangent vector is the 0 vector because of its chart. $\endgroup$
    – jizhou
    May 5 '16 at 14:31
  • $\begingroup$ This is what I would call the differentiable (or smooth) structure (i.e. a hausdorff and second countable topological space together with a $C^{\infty}$ atlas). But the tangential map (and the immersion property) does not depend on the choice of compatible(!) chart. And this is, what is usually meant when speaking of independence of charts. $\endgroup$
    – hase_olaf
    May 7 '16 at 15:39

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