Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $$f_\ast(X)(i) = X(i\circ f) = X(f) = df(X).$$

share|cite|improve this answer
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? – Mr. K 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.