$d_pf$ is a linear invertible map.

Question

If $f: S_1\to S_2$ is a diffeomorphism, then $$d_pf: T_p(S_1)\to T_q(S_2)$$ is an invertible linear map and $$(d_pf)^{-1}=d_qf^{-1}$$ for any $p\in S_1$ $q\in S_2$ and $f(p)=q$

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I cannot prove this statement. Please show me its proof. Thank you.

Answer 1

I started typing this up earlier today, and after taking a break to answer a phone call, I noticed that in those few minutes, judging from the comments, the OP's concerns seem to have been resolved. But I figured I'd post it anyway; it can be thought of as fleshing out some of the details of Daniel Rust's answer. I use an approach which emphasizes computation in a local coordinate system; hopefully my readers can glean some additional insight from this answer.

Since this basically a local problem, we can choose open neighborhoods $U, V$ of $p$ and $q$, repectively, and work in those. So in $U$ we have local coordinates $x_1, x_2, . . . , x_n$ around $p$, as in $V$ do we have local coordinates $y_1, y_2, . . . , y_n$ around $q$. Representing $f$ in terms of these coordinates, we have component functions $f_i$ such that

$y_i = f_i(x_1, x_2, . . . , x_n), \tag{1}$

and likewise the components of $f^{-1}$ with

$x_j = f_j^{-1}(y_1, y_2, . . . , y_n). \tag{2}$

If we now substitute the equations (1) into (2), we obtain relations of the form

$x_j = f_j^{-1}(f_1(x_i), f_2(x_i), . . . , f_n(x_i)), \tag{3}$

where I have typed $f_k(x_i)$ as a shorthand form for $f_k(x_1, x_2, . . . , x_n)$, i.e.,

$f_k(x_i) = f_k(x_1, x_2, . . . , x_n), \tag{4}$

and so forth. If we now differentiate (3) with respect to $x_l$ using the chain rule, we see that

$\partial x_j / \partial x_l = \sum_{k = 1}^{k = n} (\partial f_j^{-1} / \partial y_k)(\partial f_k / \partial x_l), \tag{5}$

in which the expression $\partial f_j^{-1} / \partial y_k$ is evaluated in accord with (1), and $\partial f_k / \partial x_l$ is evaluated at $x_1, x_2, . . . , x_n$; but

$\partial x_j / \partial x_l = \delta_{jl}, \tag{6}$

so that

$\sum_{k = 1}^{k = n} (\partial f_j^{-1} / \partial y_k)(\partial f_k / \partial x_l) = \delta_{jl} , \tag{7}$

which may be interpreted as showing the product of the matrices $\partial f_j^{-1} / \partial y_k$ and $\partial f_k / \partial x_l$ is the identity matrix. In a similar manner, by applying the same procedure to the equations

$y_j = f_j(x_1, x_2, . . . , x_n) = f_j(f_k^{-1}(y_l)) \tag{8}$

it may be seen that

$\sum_{k = 1}^{k = n}(\partial f_j / \partial x_k)(\partial f_k^{-1} / \partial y_l) = \partial y_j / \partial y_l = \delta_{jl}; \tag{9}$

taken together, (7) and (9) show that the matrces $[\partial f_j / \partial x_k]$ and $[(\partial f_k^{-1} / \partial y_l]$ are each both left and right inverses of one another. Since the matrix $[\partial f_j / \partial x_k]_p$ is in fact the coordinate representaion of the linear map $d_pf: T_p(S_1)\to T_q(S_2)$, as is $[\partial f_k^{-1} / \partial y_l]_q$ that of $d_qf^{-1}:T_q(S_2) \to T_p(S_1)$, we see that (7) and (9) are in fact the coordinate expressions for the equations $(d_qf^{-1})(d_pf) = I$ and $(d_pf) (d_qf^{-1}) = I$; thus we have shown that $d_pf$ is in fact invertible and that $(d_pf)^{-1}=d_qf^{-1}$ as was required. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 2

As $f$ is a diffeomorphism, it has a smooth inverse $f^{-1}\colon S_2\to S_1$. Using the chain rule we see that $$d_p\mbox{Id}_{S_1}=d_p(f^{-1}\circ f)=d_{f(p)}f^{-1}\circ d_pf.$$ Can you see how this implies that $d_pf$ is invertible and that further, $d_{f(p)}f^{-1}=(d_pf)^{-1}$?