Tangent Cotangent and the Differential

Reflexivity holds because each function agrees with itself on its whole domain, and symmetry holds because equality of functions is symmetric.

If $(U,f)$ agrees with $(V,g)$ on a neighbourhood $W_1$ of $p$ and $(V,g)$ agrees with $(Z,h)$ on a neighbourhood $W_2$ of $p$, then all three agree on the neighbourhood $W_1\cap W_2$ of $p$, so transitivity holds.

Therefore the relation is reflexive, symmetric, and transitive, hence an equivalence relation.

By [F1], there are neighbourhoods of $p$ on which $f=f^{\mathrm{\prime }}$ and $g=g^{\mathrm{\prime }}$. On the intersection neighbourhood, one also has $f+g=f^{\mathrm{\prime }}+g^{\mathrm{\prime }}$, $fg=f^{\mathrm{\prime }}{g}^{\mathrm{\prime }}$, and $cf=c{f}^{\mathrm{\prime }}$.

Therefore the germs determined by these sums, products, and scalar multiples are the same, so the operations in [F2] are representative independent.

Hence the algebra operations are well defined on germs.

Let $1$ denote the constant germ with value $1$. Applying the Leibniz rule from [F1] to $1\cdot 1$ gives $v(1)=1v(1)+1v(1)=2v(1)$, so $v(1)=0$.

Every constant germ is $c\cdot 1$ for some $c\in \mathbb{R}$, so linearity from [F1] gives $v(c\cdot 1)=cv(1)=0$.

Thus every constant germ is annihilated by $v$.

If $[f]=[g]$, then $f\circ x^{-1}$ and $g\circ x^{-1}$ agree on a neighbourhood of $a$, so their $i$th partial derivatives at $a$ are equal; hence ${\mathrm{\partial }}_i{\mathrm{\mid }}_p$ is well defined by [F1] and [F2].

Linearity is immediate from linearity of partial differentiation, and the usual product rule for partial derivatives gives ${\mathrm{\partial }}_i{\mathrm{\mid }}_p([fg])=f(p){\mathrm{\partial }}_i{\mathrm{\mid }}_p([g])+g(p){\mathrm{\partial }}_i{\mathrm{\mid }}_p([f])$.

Thus ${\mathrm{\partial }}_i{\mathrm{\mid }}_p$ satisfies [F3], so it is a derivation at $p$.

Shrink $W$ so that the segment $a+t(x-a)$ stays in $W$ for all $x\in W$ and $t\in [0,1]$. Define $g_i(x):={\int }_0^1{\mathrm{\partial }}_{i}f(a+t(x-a))dt$.

Applying the one-variable Newton-Leibniz formula from [L1] to $t\mapsto f(a+t(x-a))$ gives $f(x)-f(a)={\sum }_i(x^i-a^i)g_i(x)$.

Differentiation under the integral sign in [L1] shows each $g_i$ is smooth, and evaluating at $x=a$ gives $g_i(a)={\mathrm{\partial }}_{i}f(a)$.

By [L1], the coordinate operators belong to $T_{p}M$. If $v\in T_{p}M$ and $f$ is represented in the chart by $\tilde{f}$, then [L3] gives $\tilde{f}(u)-\tilde{f}(a)={\sum }_i(u^i-a^i)g_i(u)$ near $a=x(p)$; applying $v$ to the corresponding germ and using [L2], one obtains $v([f])={\sum }_{i}v([x^i])g_i(a)={\sum }_{i}v([x^i]){\mathrm{\partial }}_i{\mathrm{\mid }}_p([f])$.

Step 1.1 shows $v={\sum }_{i}v([x^i]){\mathrm{\partial }}_i{\mathrm{\mid }}_p$, so the coordinate derivations span $T_{p}M$.

If ${\sum }_i{c}_i{\mathrm{\partial }}_i{\mathrm{\mid }}_p=0$, apply this derivation to the coordinate germ $[x^j]$; only the $j$th term survives, so $c_j=0$. Hence the coordinate derivations are linearly independent.

Therefore ${\mathrm{\partial }}_1{\mathrm{\mid }}_p,\dots ,{\mathrm{\partial }}_n{\mathrm{\mid }}_p$ form a basis of $T_{p}M$.

Choose a smooth chart around $p$; its coordinate basis has exactly $n$ vectors.

By [L1], those $n$ coordinate derivations form a basis of $T_{p}M$.

Therefore $\dim T_{p}M=n$.

Both sides are derivations by [L1], so it is enough to compare their values on the coordinate germs $[y^k]$, which form a separating family in the chart $y$.

Applying the left-hand side to $[y^k]$ gives $\mathrm{\partial }(y^k\circ x^{-1})\mathrm{/}\mathrm{\partial }{x}^i$ at $x(p)$ by definition. Applying the right-hand side to $[y^k]$ gives ${\sum }_j(\mathrm{\partial }(y^j\circ x^{-1})\mathrm{/}\mathrm{\partial }{x}^i)(x(p)){\mathrm{\partial }}_{y^j}{\mathrm{\mid }}_p([y^k])$, and only the $j=k$ term survives because ${\mathrm{\partial }}_{y^j}{\mathrm{\mid }}_p([y^k])={\delta }_{jk}$. So the two sides agree on every coordinate germ $[y^k]$.

Therefore the two derivations agree on a basis of germs and hence are equal.

Suppose $(x\circ {\gamma }_1{)}^{\mathrm{\prime }}(0)=(x\circ {\gamma }_2{)}^{\mathrm{\prime }}(0)$ in some chart $(U,x)$ around $p$, and let $(V,y)$ be any other chart around $p$. Write $a:=x(p)$ and ${\alpha }_r:=x\circ {\gamma }_r$. For each coordinate function $h^j:=y^j\circ x^{-1}$, [L1] gives smooth functions $g_1^j,\dots ,g_n^j$ near $a$ such that $h^j(u)-h^j(a)={\sum }_{i=1}^n(u^i-a^i)g_i^j(u)$ and $g_i^j(a)={\mathrm{\partial }}_i{h}^j(a)$. Substituting $u={\alpha }_r(t)$, dividing by $t$, and letting $t\to 0$ shows $(y^j\circ {\gamma }_r{)}^{\mathrm{\prime }}(0)={\sum }_{i=1}^n{g}_i^j(a)({\alpha }_r^i{)}^{\mathrm{\prime }}(0)\mathrm{.}$ Because the vectors ${\alpha }_1^{\mathrm{\prime }}(0)$ and ${\alpha }_2^{\mathrm{\prime }}(0)$ are equal, the right-hand sides agree for $r=1,2$. Hence $(y\circ {\gamma }_1{)}^{\mathrm{\prime }}(0)=(y\circ {\gamma }_2{)}^{\mathrm{\prime }}(0)$, so the relation is chart independent.

Reflexivity and symmetry are immediate from the defining equality in [F1], and transitivity holds because equality of coordinate velocity vectors in any chart is transitive.

Hence contact equivalence is a chart-independent equivalence relation.

Choose a chart $(U,x)$ witnessing contact equivalence, write $a:=x(p)$ and ${\alpha }_r:=x\circ {\gamma }_r$, and represent the germ $[f]$ by the smooth function $\tilde{f}:=f\circ x^{-1}$ near $a$. By [L1], after shrinking if needed there are smooth functions $g_1,\dots ,g_n$ near $a$ such that $\tilde{f}(u)-\tilde{f}(a)={\sum }_{i=1}^n(u^i-a^i)g_i(u)$ and $g_i(a)={\mathrm{\partial }}_i\tilde{f}(a)$.

Substituting $u={\alpha }_r(t)$ into the identity from step 1.1, dividing by $t$, and letting $t\to 0$ gives $(f\circ {\gamma }_r{)}^{\mathrm{\prime }}(0)=(\tilde{f}\circ {\alpha }_r{)}^{\mathrm{\prime }}(0)={\sum }_{i=1}^n{g}_i(a)({\alpha }_r^i{)}^{\mathrm{\prime }}(0)\mathrm{.}$ Because contact equivalence means ${\alpha }_1^{\mathrm{\prime }}(0)={\alpha }_2^{\mathrm{\prime }}(0)$ by [F2], these derivatives are equal for $r=1,2$. Thus $(f\circ {\gamma }_1{)}^{\mathrm{\prime }}(0)=(f\circ {\gamma }_2{)}^{\mathrm{\prime }}(0)$.

Therefore the two curves determine the same velocity derivation.

By [L1], sending a contact class $[\gamma ]$ to its velocity derivation $\dot{\gamma }(0)$ defines a map from curve classes to $T_{p}M$.

Choose a chart $(U,x)$ around $p$. If $v={\sum }_i{a}_i{\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p\in T_{p}M$, define a curve in the chart by $t\mapsto x(p)+t(a_1,\dots ,a_n)$ and transport it back by $x^{-1}$. Its contact class maps to $v$, so the map of step 1.1 is surjective by [L2].

If two curve classes have the same velocity derivation, then they have the same values on each coordinate germ $[x^i]$, so their coordinate velocity vectors agree; hence the classes are equal. Thus the map is injective.

The map of step 1.1 is therefore a canonical bijection between curve contact classes and derivation tangent vectors.

By [L1], the vector $v$ corresponds to some contact class of smooth curves through $p$.

Any representative of that contact class has velocity derivation equal to $v$.

Therefore $v$ is the velocity of a smooth curve through $p$.

If $[g]=[h]$ at $F(p)$, then $g$ and $h$ agree on some neighbourhood $V$ of $F(p)$. By continuity of $F$, the inverse image of $V$ is a neighbourhood of $p$ on which $g\circ F=h\circ F$.

Therefore $g\circ F$ and $h\circ F$ define the same germ at $p$, so the pullback germ in [F2] is well defined.

Hence the target germ determines a unique source germ under pullback by $F$.

By [F2], the formula of [F1] is well defined on target germs.

If $v\in T_{p}M$, then $d{F}_p(v)$ is linear because $v$ is, and it satisfies the Leibniz rule because $v([gh]\circ F)=v((g\circ F)(h\circ F))$ and [F3] applies.

The assignment $v\mapsto d{F}_p(v)$ is linear because the defining formula of [F1] is linear in $v$. Therefore $d{F}_p$ is a well-defined linear map into $T_{F(p)}N$.

By [L1], it is enough to compute the values of $d{F}_p({\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p)$ on the coordinate germs $[y^j]$ at $F(p)$.

By [F1], one has $d{F}_p({\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p)([y^j])={\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p([y^j\circ F])=\mathrm{\partial }{\tilde{F}}^j\mathrm{/}\mathrm{\partial }{x}^i$ at $x(p)$.

The displayed linear combination of the ${\mathrm{\partial }}_{y^j}{\mathrm{\mid }}_{F(p)}$ has exactly the same values on all coordinate germs $[y^j]$, so it equals $d{F}_p({\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p)$.

Let $[g]$ be a smooth germ at $F(p)$. By [F1], $d{F}_p(\dot{\gamma }(0))([g])=\dot{\gamma }(0)([g\circ F])$.

By [F2], the right-hand side is $((g\circ F)\circ \gamma {)}^{\mathrm{\prime }}(0)=(g\circ (F\circ \gamma ){)}^{\mathrm{\prime }}(0)$, which is exactly the value of the velocity derivation of $F\circ \gamma$ on $[g]$.

Since the two derivations agree on every germ $[g]$, they are equal.

Let $v\in T_{p}M$ and let $[h]$ be a smooth germ at $G(F(p))$. Then $d(G\circ F{)}_p(v)([h])=v([h\circ G\circ F])$ by [F1].

Also $(d{G}_{F(p)}\circ d{F}_p)(v)([h])=d{F}_p(v)([h\circ G])=v([h\circ G\circ F])$ by two uses of [F1].

The two linear maps agree on every $v$ and every $[h]$, so they are equal.

Apply [L1] to $F^{-1}\circ F={\mathrm{id}}_M$ and $F\circ F^{-1}={\mathrm{id}}_N$.

The resulting identities show that $d(F^{-1}{)}_{F(p)}$ is both a left and right inverse of $d{F}_p$.

Hence $d{F}_p$ is a linear isomorphism.

By [L1], every tangent vector has the form $v={\sum }_{i}v([x^i]){\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p$.

Applying [F1] to such a vector gives $d{f}_p(v)={\sum }_{i}v([x^i]){\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p([f])={\sum }_{i}v([x^i])\mathrm{\partial }(f\circ x^{-1})\mathrm{/}\mathrm{\partial }{x}^i$ at $x(p)$.

This is exactly the action of the covector ${\sum }_i(\mathrm{\partial }(f\circ x^{-1})\mathrm{/}\mathrm{\partial }{x}^i)(x(p))d{x}_p^i$ on every $v$, so the two covectors are equal.

Let $C$ be a connected component of $M$ and fix $p\in C$. We claim that the fiber $S:=\{q\in C:f(q)=f(p)\}$ is open in $C$.

Let $q\in C$. Choose a smooth chart $(U,x)$ around $q$ and an open Euclidean ball $B$ with $x(q)\in B\subseteq x(U)$; put $W:=x^{-1}(B)$. For any $r\in W$, the map ${\gamma }_{q,r}(t):=x^{-1}((1-t)x(q)+tx(r))(0\le t\le 1)$ is a smooth curve in $W$ from $q$ to $r$ by [F1]. For each $t_0\in [0,1]$, apply [L1] to the shifted curve $s\mapsto {\gamma }_{q,r}(t_0+s)$ at $s=0$; since $d{f}_{{\gamma }_{q,r}(t_0)}=0$, this gives $(f\circ {\gamma }_{q,r}{)}^{\mathrm{\prime }}(t_0)=0$. Thus [L2] makes $f\circ {\gamma }_{q,r}$ constant on $[0,1]$, so $f(r)=f(q)$. Therefore $f$ is constant on $W$.

Step 1.2 shows that every fiber of $f{\mathrm{\mid }}_C$ is open in $C$. Hence $S$ is open, and so is its complement $C\setminus S$, which is the union of the other fibers. Since $C$ is connected and $S$ is nonempty, one must have $C\setminus S=\mathrm{\varnothing }$. Therefore $f$ is constant on $C$.

Because the connected component $C$ was arbitrary, $f$ is constant on each connected component of $M$.

If $v={\sum }_i{v}^i{\mathrm{\partial }}_{x^i}{\mathrm{\mid }}_p$, then [L1] gives $v={\sum }_j{w}^j{\mathrm{\partial }}_{y^j}{\mathrm{\mid }}_p$ with $w=J(a)v$, where $a:=x(p)$ and $J(a)=D(y\circ x^{-1})(a)$. Hence $\tilde{y}\circ {\tilde{x}}^{-1}(a,v)=(y\circ x^{-1}(a),J(a)v)$.

The base part $y\circ x^{-1}$ is smooth, the matrix-valued map $a\mapsto J(a)$ is smooth, and matrix-vector multiplication is polynomial in the entries; therefore the transition map is smooth.

Reversing the roles of $x$ and $y$ gives the inverse transition, whose fiber matrix is $J(a{)}^{-1}$. The smoothness of this inverse matrix field follows from [L2], so the inverse transition is smooth.

For each induced bundle chart $\tilde{x}:{\pi }^{-1}(U)\to x(U)\times {\mathbb{R}}^n$, declare the sets ${\tilde{x}}^{-1}(O)$, with $O$ open, to be basic open sets. The transition homeomorphisms from [L1] make these families agree on overlaps, so they define a topology on $TM$ for which every $\tilde{x}$ is a homeomorphism onto the open set $x(U)\times {\mathbb{R}}^n\subseteq {\mathbb{R}}^{2n}$.