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Let M and N be smooth manifolds: $$\mathbb{R}\xleftarrow{g} M \xrightarrow{h}N\xrightarrow{f}\mathbb{R}$$ and of course: $$TM\xrightarrow{h_*}TN$$ I need to show that: $$h_{*}(gX)f=(g\circ h^{-1})h_{*}Xf$$ where $X\in TM$. Tried by substituting $f\rightarrow g\circ h^{-1}$ but it doesn't make sens. Any suggestions?

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  • $\begingroup$ Welcome to MSE! Good job posting with the formatting! $\endgroup$ – user370967 Jun 5 '17 at 9:04
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Substitute $Y = gX$. The pushforward then becomes $(h_*Y)_q(f) = Y_{h^{-1}(q)}(f\circ h)$. Then you can reenter your substitution to obtain: $g(h^{-1}(q))X_{h^{-1}(q)}(f\circ h)$ or equivalently $(g\circ h^{-1})h_*X(f)$.

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  • $\begingroup$ How is (f o h^-1) even possible? $\endgroup$ – kishony Jun 5 '17 at 9:33
  • $\begingroup$ My apologies for this. I corrected my mistake. $\endgroup$ – NDewolf Jun 5 '17 at 9:39
  • $\begingroup$ Well.... that was easy. Thank you! $\endgroup$ – kishony Jun 5 '17 at 9:57

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