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Let $X_1, X_2$ be vector fields over a smooth manifold $M$ and likewise $Y_1, Y_2$ vector fields over a smooth manifold $N$. Furthermore, let $X_1$ and $Y_1$ aswell as $X_2$ and $Y_2$ be $\Phi$-related for some morphism $\Phi: T M \to T N$ (i.e. we have $\Phi \circ Y_i = X_i \circ \phi$ where $\phi: M \to N$ is the projection mapping of $\Phi$). I now want to show that the commutator/Lie bracket $[X_1, Y_1]$ is also $\Phi$-related to $[X_2, Y_2]$.

Now I tried to go by the definition of the commutator and write out

$$[X_1, Y_1] \circ \phi = X_1 Y_1 \circ \phi - Y_1 X_1 \circ \phi \\ = X_1 \circ (\Phi \circ Y_2) - Y_1 \circ (\Phi \circ X_2) \\$$

But I'm not really sure how to go from there as I don't think I could interchange $X_i \circ \Phi$ for $\Phi \circ X_i$? I would need to show that this expression is equal to

$$\Phi \circ [X_2, Y_2] = \Phi(X_2 Y_2 - Y_2 X_2)$$

but approaching it from this side, I again don't know how to continue from here, as I don't think I could just "pull in" $\Phi$ into the individual summands? Is there something about the commutator that I'm misunderstanding or forgetting? It seems like an easy thing to show yet I'm a bit lost here.

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The correct statement of the problem is that if $\phi : M \rightarrow N$ are smooth map and $X_i \in \mathfrak{X}(M)$ and $Y_i \in \mathfrak{X}(N)$ are $\phi-$related, then the commutator $[X_1,X_2]$ is also $\phi-$related to $[Y_1,Y_2]$. To show that, let $f \in C^{\infty}(N)$ be any smooth function on $N$, by hypothesis verify that $$ X_1X_2 (f \circ \phi) = \dots = (Y_1Y_2 f) \circ \phi $$ and similarly $X_2X_1 (f \circ \phi) = \dots = (Y_2Y_1 f) \circ \phi$. And then using definition of Lie bracket to show $$ [X_1,X_2](f \circ \phi) = ([Y_1,Y_2]f) \circ \phi. $$

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  • $\begingroup$ Thank you, for the correction (wasn't aware of the mistake in the statement) aswell as for the answer. Could you maybe explain what happens during the "$\dots$"-part? I'm not sure if I can follow there. $\endgroup$ – moran Nov 19 '17 at 21:32
  • $\begingroup$ Just use the fact that $X_i$ and $Y_i$ are related. That is $X_2(f\circ \phi) = (Y_2 f) \circ \phi$ . So $$ X_1X_2 (f \circ \phi) = X_1 ((Y_2f) \circ \phi) $$ Now by regard $Y_2 f$ as a function in $N$, apply the hypothesis that $X_1$ related to $Y_1$ as above to get $X_1X_2(f \circ \phi) = X_1 ((Y_2f) \circ \phi) = (Y_1Y_2f)\circ \phi$. $\endgroup$ – Sou Nov 19 '17 at 21:40

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