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.
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 $H^1(O_C(D))=H^0(O_C(KD))=0$ because $\deg KD = \deg D <0 $ since on elliptic curve $\deg K=0$. 


In general, for any nonconstant morphism $f:C \to \mathbb P^n$, from a $1$dimensional CohenMacaulay (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 pullback 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 nonconstant, this is an antiample line bundle on $C$ and hence has no global sections. Remark: One needs the CohenMacaulay condition for Serre duality and so that $\omega_C$ is sensible. 

