Q:

Flows of $f$-related vector fields

If we have a smooth map between manifolds $f : M \rightarrow N$ and (say, complete) vector fields $X$ in $M$ and $Y$ in $N$, we say they are $f$-related when $Y(f(x)) = d_x f(X(x))$ for all $x \in M$. I've heard it then claimed that, if $X$ is $f$-related to $Y$, their flows will satisfy the relationship: $f(\phi_{X}^{t}(x)) = \phi_{Y}^{t}(f(x))$.
Can anyone indicate an argument or reference for this? I was told to simply differentiate that relationship, but then it seems we get $df(X(\phi_{X}^{t}(x))) = Y(\phi_{Y}^{t}(f(x)))$; we don't know if those are equal, and even if they were I'm not sure how that implies anything.

A:

Suppose $\gamma: \mathbb{R} \rightarrow M$ is an integral curve for $X$. That is, $d\gamma_t(\partial_t) = X_{\gamma(t)}$.
Consider $f \circ \gamma: \mathbb{R} \rightarrow M$. Then $d(f\circ\gamma)_t(\partial_t) = df_{\gamma(t)} \circ d\gamma_t(\partial_t) = df_{\gamma(t)} (X_{\gamma(t)}) = Y_{f(\gamma(t))} = Y_{(f\circ\gamma)(t)}$. So $f\circ \gamma$ is an integral curve for $Y$.
What is the flow $\phi^t_X$? Given $x \in X$, find an integral curve $\gamma$ with $\gamma(0) = x$. Then $\phi^t_X(x) = \gamma(t)$. (This is either the definition, or follows from uniqueness for solutions to ODE's, depending on how you set things up.)
But then $f\circ \gamma$ is an integral curve with $f\circ\gamma(0) = f(x)$. Thus $\phi^t_Y(f(x)) = (f\circ\gamma)(t) = f \circ \phi^t_X(x)$.

