Proof of Darboux Theorem I'm reading the proof of Darboux Theorem from Spivak's book on Mechanics. 
After he gets a 1-form $\alpha$ and the form $\omega_t=\omega+t(\omega_1-\omega)$ he defines $X^t$ as a field with $X^t(0)=0$ and $i_{X^t}\omega_t=-\alpha$ for $t \in [0,1]$.
Then he defines the 1-parameter family $f_t$ of diffeomorphisms generated by $X^t$. This is what I don't understand. Shouldn't every $X^t$ generate a whole 1-parameter family of diffeomorphisms and not just one? Which diffeomorphism is $f_t$?
 A: All that's going on is the vector field is varying as you're integrating it. We don't separately take integral curves for each. Instead we make different integral curves which follow the vector field $X^t$ at time $t$. Details:
Compare: Time-independent
A $1$-parameter family $f$ of diffeomorphisms generated by a single (i.e. time-independent) vector field $X$ comprises integral curves $\gamma$ satisfying $$\frac{d\gamma}{dt}(t) = X_{\gamma(t)}$$ That is, the derivative of $\gamma$ at $t$ equals the value of the vector field $X$ at the point $\gamma(t)$.
Time-dependent
When we allow the vector field to vary with time, so we have a smoothly varying family of vector fields with one for each $t$, called $X^t$, then we ask that the integral curves satisfy $$\frac{d\gamma}{dt}(t) = X^t_{\gamma(t)}$$ That is, the derivative of $\gamma$ at $t$ equals the value of the vector field $X^t$ at $\gamma(t)$.

In case it helps, in coordinates on $\mathbb{R}^n$, this is the difference between having a vector field $V: \mathbb{R}^n \rightarrow \mathbb{R}^n$ and solving $$\gamma'(t) = V(\gamma(t))$$ and having a time-dependent vector field $V: \mathbb{R} \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ and solving $$\gamma'(t) = V(t, \gamma(t))$$
A: $f_t$ is the diffeomorphism that takes a point $P$ to $U$, where $U$ is the point on the integral curve $\gamma$ of $X$ through $P$ with $\gamma(0) = P$ and $\gamma(t) = U$. 
In words: $f_t$ is the result of flowing along $X$ for time $t$. 
