I have a differential map on manifolds $\psi: M_1 \to M_2$, where $\phi_1$ and $\phi_2$ are one parameter group of transformations on $M_1$ and $M_2$ respectively.Now if $\phi_{2t} \circ \psi = \psi \circ \phi_{1t} $, then I need to show X and Y which are infinitesimal generators of $\phi_1$ and $\phi_2$ are $\psi$-related.
I attempted to solve in the following way
Two vector fields are $\psi$-related if $Y_{\psi(x)}= \psi_{*x} X_x$
so $Y_{\psi(x)}= \psi_{*x} (\phi_{1t})'_0$
Here I run into problem as I am having trouble in using the definition of push forward. Please someone suggest how should I proceed.(This is from Differentiable Manifolds:A Theoretical Physics Approach by Gerardo F. Torres del Castillo problem 2.8)
The required statement is simply a statement of the chain rule for $Y_{\psi(x)}$. On the LHS of $\phi_{2t} \circ \psi = \psi \circ \phi_{1t}$, $\psi$ simply takes $x$ to $\psi(x)$ and the tangent vector is calculated there as $Y(\psi(f(x))) = \mathrm{d}_t \phi_{2t}(\psi(f(x)))$. On the RHS, you calculate the tangent vector with the chain rule: in $T_x M_1$ the tangent vector is $X(f)=\mathrm{d}_t \phi_{2t}(f(x))$ and then it gets linearly mapped by the pushforward (the Jacobi matrix) $\mathrm{d} \psi$ to $T_{\psi(x)} M_2$. Equate these two and you're there: $\mathrm{d} \psi\,X_x = Y_{\psi(x)}$.
To get this, don't rush it. You need to mull over the definition of the pushforward for a while. Draw a commutative diagram. The notations can seem highly dense and arcane at first and you need practice at simply applying them, which is exactly the point of this question.