Identifications of functions between $C^\infty(M)$ and $Vec(M)$ with $C^\infty(T^*M)$. We have the following two identifications:
(i) The identification of smooth functions on a manifold $M$ with functions that are constant on fibers:$$C^\infty(M) \cong C^\infty_{\text{const}}M,$$via $\alpha \mapsto \pi^*\alpha = \alpha \circ \pi$, where $\pi$ is the canonical projection $\pi : T^*M \rightarrow M$.
(ii) The identification of smooth vector fields with smooth functions that are linear on fibers:$$Vec(M) \cong C^\infty_{\text{lin}}M,$$via $X \mapsto a_X$ where $a_X(\lambda) = \langle \lambda, X(q)\rangle$ where $\pi(\lambda) = q$.
Question: I know these identifications must be based on linear algebra but I haven't been able to really understand why these happen so naturally. I would appreciate if anyone could explain both (i) and (ii) with the language of linear algebra.
 A: *

*This is essentially a basic set-theoretic thing, not specific to manifolds. Suppose $X,Y,Z$ are sets, $\pi:X\to Y$ is a surjective function. Denote $\mathcal{F}(Y,Z)$ to be the set of all functions $f:Y\to Z$, and let $\mathcal{F}_{\pi}(X,Z)$ be the set of all functions $g:X\to Z$ which are constant on the fibers of $\pi$ (i.e $\pi(x_1)=\pi(x_2)$ implies $g(x_1)=g(x_2)$). Then, the mapping $\mathcal{F}(Y,Z)\to \mathcal{F}_{\pi}(X,Z)$, $f\mapsto f\circ \pi$ is a bijection. This is just saying that if $f:Y\to Z$ is any function, then $f\circ \pi:X\to Z$ is a function which is constant on the fibers of $\pi$ (this is trivial). Conversely, if $g:X\to Z$ is a function which is constant on the fibers of $\pi$, then we can define $f:Y\to Z$ by setting $f(y)$ to be the constant value of $g$ when restricted to the fiber $\pi^{-1}(\{y\})$; and the thus constructed $f$ satisfies $g=f\circ \pi$. If you’ve learnt about quotient spaces (either for groups/vector spaces/rings/fields or topology etc) then this should be a familiar concept (at this point you can draw a nice triangular commutative diagram for $f,g,\pi$).

The only additional thing, in the case of smooth manifolds, you have to verify is the smoothness of maps involved. You have to show that if $f:Y\to Z$ and $\pi:X\to Y$ are smooth then $f\circ \pi:X\to Z$ is smooth (trivial from the chain rule). Conversely, you need to show that if $\pi:X\to Y$ is a surjective submersion (which the projection of bundles certainly is) and $g:X\to Z$ is a smooth map then the unique map $f:Y\to Z$ satisfying $g=f\circ \pi$ (whose existence and uniqueness we know from our previous set-theoretic discussion) is also a smooth map. Here, you need to use the local canonical form for submersions.


*A vector field on $M$ is a mapping $X:M\to TM$ such that $\pi_{TM}\circ X=\text{id}_M$. So, for each $p\in M$, we have an element $X_p\in T_pM$. From linear algebra, you know about double duality (for every finite-dimensional vector space $V$, the mapping $\theta:V\to V^{**}$, defined by evaluation, $\theta(v)(\lambda):=\lambda(v)$, is an isomorphism), so $X_p\in T_pM\cong(T_pM)^{**}$. Being in the double dual means you can identify $X_p$ with a linear map $T_p^*M\subset T^*M\to\Bbb{R}$, i.e you have a map defined on a subset of $T^*M$, but not just any arbitrary subset; it is defined on the entire fiber, which is a vector space, and the map thus defined is linear on the fiber. If you now do this for each point $p\in M$, then you can stitch these maps defined on the various $T_pM^*$’s together to get a new map $f:T^*M\to\Bbb{R}$, which by construction is fiber-wise linear.

I’ll leave it to you to check smoothness is preserved.
