Can every linear map between tangent spaces be realized as a differential of a map? Let $M,N$ be smooth manifolds, $p \in M,q \in N$. Suppose $T \in \text{Hom}(T_pM,T_qN)$. Does there exists a smooth map $f:M \to N$ such that $T=df_p$?
I am asking about global realization; The local problem is trivial (i.e when $M=\mathbb{R}^m,N=\mathbb{R}^m$ just take an affine map).
The problem with passing to the global setting is that in general, not every local map can be extended to a global one.
In the case where $N=\mathbb{R}^k$, we can extend any smooth map defined on a closed subset of $M$ to a function in $C^{\infty}(M,N)$. (This is lemma 2.26 in Lee's book on smooth manifolds).
A general extension (for arbitrary codomains) does not exist even for continuous maps, due to topological reasons.
 A: As you said, the problem can be solved locally. So let $p\in U$ be a neighborhood of $p$ where such map $f$ can be defined.
Now consider contractible open sets $p\in A\subset B\subset U$ and modify the original map as follows.
On $A$ you leave it unchanged.
On $B\setminus A$ you modify $f$ smoothly so that $f$ becomes constant on the boundary of $B$.
Finally you define $f$ on $M\setminus B$ to be constant as well.
A: Extending from a small ball is much easier than extending from an arbitrary closed subset. What follows is essentially an explicit description of AdLibitum's answer. 
Choose charts $(U,x^i)$ about $p$, $(V,y^\alpha)$ about $q$ and let $\eta$ be a bump function that is $1$ in some neighbourhood of $p$ and compactly supported inside $U$; and define in coordinates $$y^\alpha(f(x)) = \eta(x) T^\alpha_i x^i.$$ By shrinking the support of $\eta$ if necessary you can ensure that the RHS stays in $V$, so this makes sense. Since $f(x)=q$ for all $x \in U \setminus\mathrm{supp}\;\eta$, extending it with the value $q$ everywhere else gives you a smooth global map.
