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The pushforward of a map $F:M \to N$ at a point $P \in M$ is defined as $F_*:T_P(M) \to T_{F(P)}(N)$ where $$(F_*X)(f) = X(f \circ F)$$

where $X \in T_P(M)$.

The differential of a function $f$ defined on $M$ at $P$ is $$df_P(X_P) = X_Pf.$$

What is the relation? How to show that the differential is got from the pushforward definition? Presumable for the differential case, $f:M \to \mathbb{R}$, but I can't get the answer.

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Where do you see the difference between differential and pushforward in your definition? Because I don't see one. – Nils Matthes Sep 10 '12 at 11:20
up vote 9 down vote accepted

The confusion can be resolved by understanding the diffeomorphism $T_q\mathbb{R}\cong\mathbb{R}$: every derivation $C^\infty(\mathbb{R})\to\mathbb{R}$ based at the point $q\in\mathbb{R}$ comes in the form $t\,\frac{\partial}{\partial x}|_q$ for some $t\in\mathbb{R}$. The diffeomorphism $T_q\mathbb{R}\to\mathbb{R}$ can therefore be thought of as "evaluate the identity $i:x\mapsto x$".

Let $f:M\to\mathbb{R}$ be smooth. Thinking of $f$ as a map of manifolds, its derivative $f_\ast:T_p M\to T_q \mathbb{R}$ is defined by $f_\ast(X)(g) = X(g\circ f)$, just as you say. Now observe $$f_\ast(X)(i) = X(i\circ f) = X(f) = df(X).$$

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Nice answer. If you use $t$ as the coordinate on $\mathbb{R}$ this identity amounts to setting $d/dt=1$. It's the same identification that makes one-dimensional physics both easier and tricker due to sign-conventions breaking down. – James S. Cook Sep 10 '12 at 13:45
I just don't understand why the identity comes up. We could use any map $f:\mathbb{R}\to\mathbb{R}$ instead and then we wouldn't reduce the pushforward to the differential. What am I missing? – Iberê Kuntz Apr 16 '15 at 0:35
@IberêKuntz You can certainly use another map $g$ (instead of $f$, to reduce notation clash) for the identification $T_q\mathbb{R}\cong \mathbb{R}$ if you like, provided that $g'(q)=1$. Then you'll find that, at $q=f(p)$, $f_*(X)(g) = g'(q) X(f) = X(f)$. – Sean Eberhard Apr 16 '15 at 8:32

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