Question: If $C$ is a smooth elliptic curve and $f: C \to \mathbb P^n$, then 
$H^1(C,f^\*T\_{\mathbb P^n}) = 0.$ How do I prove this? The implication is that map
from $C$ to $\mathbb P^n$ is unobstructed.

Answer: 
I am a beginner but here is my attempt:

the Euler sequence on $\mathbb{P}^n$ pulls back to $0 \to O\_C \to O\_C(1)^{n+1} \to f^\*T\_{\mathbb{P}^n} \to 0$ and so from the associated long exact sequence we want to show $H^1(O\_C(1))=0$ (as $H^2(O\_C)=0$ since $\dim C = 1 <2$). Let $D$ be an effective divisor on $C$ so that $O\_C(1)=O\_C(D)$. By Serre duality we have $H^1(O\_C(D))=H^0(O\_C(K-D))=0$ because $\deg K-D = \deg -D <0 $ since on elliptic curve $\deg K=0$.
In general, for any *non-constant* morphism $f:C \to \mathbb P^n$, from a $1$-dimensional Cohen-Macaulay (for instance reduced) curve $C$, one has that $$H^1(C,f^\*T\_{\mathbb P^n}\otimes \omega\_C)=0.$$

Indeed, (as already pointed out by Al e) considering the pull-back of the Euler sequence
$$0 \to f^\*\mathscr O\_{\mathbb P^n} \to f^\*\mathscr O\_{\mathbb P^n}(1)^{\oplus (n+1)} \to f^\*T\_{\mathbb{P}^n} \to 0$$
and using the fact that $H^2(C,\mathscr O\_C)=0$ automatically by dimension considerations, 
it is enough to prove that $$H^1(C, f^\*\mathscr O\_{\mathbb P^n}(1)\otimes \omega\_C)=0.$$
By Serre duality this is dual to $H^0(C,f^\*\mathscr O\_{\mathbb P^n}(-1))$ and since $f$ is non-constant, this is an anti-ample line bundle on $C$ and hence has no global sections.

*Remark*:
One needs the Cohen-Macaulay condition for Serre duality and so that $\omega\_C$ is sensible.