Jtdylktuy

I am studying differential topology and I have some trouble understanding how coordinates are induced on the tangent space at any point.

Let $M$ be an $n$-dimensional smooth manifold, and let $p in M$. Let $(U,varphi)$ be a coordinate chart around $p$, with coordinates $x_1,dots,x_n$. The tangent space $T_p M$ is defined to be $(mathfrak{m}_p/mathfrak{m}_p^2)^*$, where $mathfrak{m}_p$ is the $Bbb{R}$-algebra of germs at $p$ that vanish at $p$.

Now, we have the following lemma (whose proof I omit).

Lemma. If $x_1,dots,x_n$ are coordinates around $p in M$ and $f in mathfrak{m}_p$, then there exist germs $g_1,dots,g_n$ at $p$ such that $f = sum_i x_i g_i$.

Using this we can show that the images of the coordinates in $mathfrak{m}_p/mathfrak{m}_p^2$ form a basis for this vector space.
Indeed, we have the following:

1. 
   $f in mathfrak{m}_p^2 iff g_i in mathfrak{m}_p$ for all $i = 1,dots,n$, since clearly $x_i in mathfrak{m}_p$ for all $i = 1,dots,n$.


2. If $f = sum x_i g_i in mathfrak{m}_p$, then $f = sum x_i (g_i - g_i(0)) + sum g_i(0) x_i implies bar{f} = sum g_i(0) bar{x}_i$. Here, the bar indicates the image in $mathfrak{m}_p/mathfrak{m}_p^2$.

Let $frac{partial}{partial x_1},dots,frac{partial}{partial x_n}$ be the dual basis of $bar{x}_1,dots,bar{x}_n$. Then, we say that the coordinates $x_1,dots,x_n$ around $p$ induce the coordinates $frac{partial}{partial x_1},dots,frac{partial}{partial x_n}$ on $T_p M$.

If my understanding so far is correct, what this means is that I can talk about the coordinates $frac{partial}{partial x_1},dots,frac{partial}{partial x_n}$ on $T_p M$ induced by $varphi$ only when the coordinates around $p$ given by $varphi$ are such that $varphi(p) = 0$. Otherwise, in the above construction it is no longer true that the coordinates $x_1,dots,x_n$ define germs at $p$ that vanish at $p$.

So, in particular this means that although $(U,varphi)$ defines coordinates around every point $p' in U$, $varphi$ does not induce coordinates on $T_{p'} M$ if $varphi(p') neq 0$.

Is this true? I would appreciate any helpful comments and answers in this regard.

Further motivation.

This line of thinking emerged while I was trying to solve the following problem.

Exercise. Let $f : M to N$ be a smooth map between smooth manifolds. Show that the set of critical points of $f$ form a closed subset of $M$.

The natural method in my opinion is to show that the complement, that is the set of regular points, is open in $M$. So if $p in M$ is a regular point, then $df_p : T_p M to T_{f(p)} N$ is of maximal rank. If suitable coordinate charts $(U,varphi)$ and $(V,psi)$ are chosen around $p$ and $f(p)$, respectively, defining coordinates $x_1,dots,x_m$ and $y_1,dots,y_n$, respectively, then $df_p$ is represented by the Jacobian of $tilde{f} equiv psi circ f circ varphi^{-1}$ at the point $varphi(p)$, with respect to the bases $frac{partial}{partial x_1},dots,frac{partial}{partial x_m}$ on $T_p M$ and $frac{partial}{partial y_1},dots,frac{partial}{partial y_n}$ on $T_{f(p)} N$.
Now, I know that for every $varphi(p')$ in a small enough neighbourhood around $varphi(p)$, $mathrm{Jac}(tilde{f})_{varphi(p')}$ is also of maximal rank.

At this point, I would like to be able to say that $mathrm{Jac}(tilde{f})_{varphi(p')}$ represents the map $df_{p'} : T_{p'} M to T_{f(p')} N$ with respect to the bases $frac{partial}{partial x_1},dots,frac{partial}{partial x_m}$ on $T_{p'} M$ and $frac{partial}{partial y_1},dots,frac{partial}{partial y_n}$ on $T_{f(p')} N$.

If I could, then I would be done. But, I went back to the definition of $frac{partial}{partial x_1},dots,frac{partial}{partial x_m}$ as I've understood it above, and I'm not so sure that I am correct in saying so.

There are several equivalent definitions of the tangent space. Consider $C_p(M)$ the $mathbb{R}$-algebra of germs at $p$ and define the tangent space of $M$ at $p$ as the dual space of the quotient $C_p(M)/(mathfrak{m}_p^2+mathbb{R})$, where the copy of $mathbb{R}$ in the denominator is the vector subspace of constant functions. With this definition, the value of $varphi(p)$ does not matter.

Let $G_p$ be the set of germs at $p$. Let $(U,varphi)$ and $(V,psi)$ be coordinate charts around $p$, with coordinates $x_1,dots,x_n$ on $U$ and $y_1,dots,y_n$ on $V$. If $f : M to Bbb{R}$ is a smooth function such that
$$
frac{partial f}{partial x_i}bigg|_{p} = 0 quad text{ for all } quad i = 1,dots,n,tag{$*$}
$$
then
$$
frac{partial f}{partial y_i}bigg|_{p} = 0 quad text{ for all } quad i = 1,dots,n.
$$
Also, if $f_1,f_2 : M to Bbb{R}$ are smooth functions that agree on a small neighbourhood around $p$, then we have
$$
frac{partial f_1}{partial x_i}bigg|_p = frac{partial f_2}{partial x_i}bigg|_p quad text{ for all } quad i = 1,dots,n.
$$
Let $S_p$ be the subspace of stationary germs at $p$, that is, those germs at $p$ for which ($*$) hold. By the above remarks, $S_p$ is a well-defined subspace of $G_p$.

Now, suppose $f in G_p$ and $x_1,dots,x_n$ are coordinates around $p$ with $(alpha_1,dots,alpha_n)$ the coordinates of $p$. Then, by the above Lemma, we have that
$$
f(x) = f(p) + sum_{i=1}^n (x_i - alpha_i) g_i(x) = f(p) - sum_{i=1}^n alpha_i g_i(p) + sum_{i=1}^n (x_i - alpha_i) (g_i(x) - g_i(p)) + sum_{i=1}^n g_i(p) x_i.
$$
The first two terms of the right-most expression are constant germs and hence stationary germs. The third term is also a stationary germ, as can be seen by differentiating and applying the chain rule. Hence, if $bar{f}$ denotes the image of $f$ in $G_p / S_p$, we have that
$$
bar{f} = sum_{i=1}^n g_i(p) bar{x}_i.
$$

So, we can still make sense of $bar{x}_1,dots,bar{x}_n$ and its dual basis $frac{partial}{partial x_1},dots,frac{partial}{partial x_n}$ when $p$ does not have coordinates $(0,dots,0)$. Note that
$G_p/S_p cong mathfrak{m}_p/(mathfrak{m}_p cap S_p)$ by the second isomorphism theorem, because $G_p = mathfrak{m}_p + S_p$. Moreover, $mathfrak{m}_p cap S_p = mathfrak{m}_p^2$. So, there is no trouble in defining the tangent space $T_p M$ to be $(G_p / S_p)^*$.

I obtained some supplementary course notes that briefly mention $T_p^* M := G_p/S_p = mathfrak{m}_p / mathfrak{m}_p cap S_p$, and $T_p M := (T_p^* M)^*$. I have only elaborated this in greater detail in my answer.

Also, the expression for $f(x)$ that I have obtained here shows that $S_p = mathfrak{m}_p^2 + Bbb{R}$, where $Bbb{R}$ denotes the subspace of constant germs. So, this answer can also be considered an elaboration of the answer by @DanteGrevino.