Why is the manifold structure on the tangent bundle unique? ...subject to the conditions that (i) the projection be smooth and that (ii) smooth sections correspond to smooth vector fields.
This homework problem is really bugging the hell out of me.  Of course it can be checked locally, so we'll look at an open neighborhood $U\in \mathbb{R}^n$ and check that $TU\cong U\times \mathbb{R}^n$ (where the isomorphism is as topological spaces with sheaves of functions that we declare to be smooth).  On the one hand, it's easy to see that $U \times \mathbb{R}^n$ satisfies these two conditions, since the projection is literally just a projection (in coordinates) and the correspondence is


*

*$X$ a vector field --> $s = (x_1,\ldots, x_n, X(\partial/\partial x_1), \ldots, X(\partial/\partial x_n))$

*$s=(s_1,\ldots, s_n)$ a smooth section --> $X = \sum_i s_i\cdot\partial/\partial x_i $.


On the other hand, I'm having a lot of trouble showing that the obvious sheaf (which I'll denote $O_{TU}$) is the unique one with properties (i) and (ii).
For example, suppose $\mathcal{F}$ is another sheaf of functions on $U\times \mathbb{R}^n$.  Suppose that $f \in \mathcal{F}(V) \backslash \mathcal{O}_{TU}$ for some open set $V\subseteq U$.  There really doesn't seem to be any way to show that this violates property (i), because what it really means is that for any $g\in O_U$, $g \circ \pi \in O_{TU}$, and the hypothesis of this statement begins with a function on $U$, not $TU$.  On the other hand, I don't see any way to make a vector field on $U$ out of $f$, and (according to this question) it feels likely that I can't necessarily get a section that detects the failure of $f$ to be $O_{TU}$-smooth.  Nor am I having much luck deriving a contradiction from $f \in O_{TU}(V) \backslash \mathcal{F}(V)$.
So, I'd welcome any suggestions as to how I should proceed.
 A: This is actually easier if you use charts. Suppose your smooth structure is given by a maximal atlas C.
Near any point $p$ in $TU$ you have a neighborhood $V=\pi^{-1} W$ for $W$ - a chart neighborhood of $\pi(p)$, so that on $V$ which there are functions $x_i \cdot\pi$ which are smooth by property (1) and functions $\frac{\partial}{\partial x_i}$, which are smooth by property (2).  But they also define a chart - a map $V$ to $R^{2n}$. So this chart has to be smooth. So we have a cover by charts smooth in both the standard atlas and the maximal atlas C. Hence the two smooth structures coincide.
Trying to do it in sheave seems a bit strange. If your structure sheaf is F, you get $2n$ functions in $F(V)$ and the fact that they define a chart should mean that you can pull back (restriction of) any function f in F to a function on part of $R^{2n}$, so that restriction is in the standard structure sheaf, and hence, (by sheaf axiom) f is too. This seems slightly fishy, even though can probably be fixed, in which case it will be the sheaf translation of the above argument.
A: First, I am pretty rusty on this stuff, so hopefully somebody
with more confidence will chime in. I started writing this as another
comment, but it didn't fit.
Anyway, I still think what I said makes sense. If
$\mathcal{E}_{\mathbb{R}^n}$ is the standard differentiable structure
on $\mathbb{R}^n$, and $\mathcal{C}_M$ is the sheaf of continuous
functions on a topological manifold $M$, then a subsheaf $\mathcal{F}$
of $\mathcal{C}_M$ is a differentiable structure on $M$ if (and only if)
1) at each point $p\in M$ there exists a neighbourhood $V$, and
  $x_1,...x_n \in \mathcal{F}(V)$ such that $q \mapsto (x_1, ...,
  x_n)$ is a homeomorphism onto an open $U \subset \mathbb{R}^n$, and
2) $f \in \mathcal{F}(V)$ iff there exists $F \in
  \mathcal{E}_{\mathbb{R}^n}(U)$, such that $f = F(x_1,...,x_n)$.
I think this definition is derived from Warner ("Foundations on
differentiable manifolds and Lie groups"), but it has paraphrased by
me and possibly been distorted through the lens of an absurdly long
time. In particular, that second condition should probably be reworked
to be more obviously a purely sheaf theoretic statement.
In any event, a proper definition of a differentiable structure is
going to resemble something like this (likewise for holomorphic
structure, or analytic structure or whatever). And the point is, that
the \emph{definition} will require that your sheaf of differentiable
functions on $M$ will be locally isomorphic to the \emph{standard}
sheaf of differentiable functions on $\mathbb{R}^n$. 
This should apply even if the topological manifold $M$ is
$\mathbb{R}^n$. This is why I say that you can't solve the problem
locally: Locally all differentiable structures are isomorphic to the
standard one on $\mathbb{R}^n$ by definition.
As to my second statement, about an "appropriate" differentiable
structure on $TM$ defining a unique differentiable structure on $M$,
this is mostly a guess. However, the qualifier "appropriate" is
important here. In particular, whatever differentiable structure on
$TM$ you choose, it has to restrict to the standard structure on
$\mathbb{R}^n$ on the fibres. In this way, I don't think there is any
choice left once you've defined the restriction on the zero section.
