Let $M,N$ be two differentiable manifolds and $f:M \rightarrow N$ be a smooth map. Define a new map $F:M\rightarrow M\times N$ by $F(p)=(p,f(p))$

I can prove first part which is F is smooth but I can not show that the second part

"Show that $F_{*}(v)=(v,f_*(v))$ where $F_*$ and $f_*$ are induced maps at a point $p$ of $M$ and $v$ is a tangent vector of $M$ at $p$"

Can anybody help?

  • 1
    $\begingroup$ Hint: think about $\pi_1 \circ F$ and $\pi_2 \circ F$ where $\pi_1 : M \times N \rightarrow M$ is the projection on the first coordiante and $\pi_2 : M \times N \rightarrow N$ the projection on the second coordinate. $\endgroup$ – Pedro Apr 19 '15 at 10:55
  • $\begingroup$ Sorry, but how does it help me? (I proved F is smooth, but I can not prove the second part) $\endgroup$ – corcia candy Apr 19 '15 at 10:57

Let us call $\pi_1 : M \times N \rightarrow M$ the projection on the first coordinate and $\pi_2 : M \times N \rightarrow N$ the projection on the second coordinate.

First of all, if $p \in M$ and $q \in N$, we need to pay attention to how the natural identification $T_{(p, q)} (M \times N) \cong T_p M \times T_q N$ works. What happens is that the map $\phi : T_{(p, q)} (M \times N) \rightarrow T_p M \times T_q N$ given by $\phi(v) = (\pi_{1 *}(v), \pi_{2 *}(v))$ is an isomorphism (this is a good exercise), and this is the isomorphism we use to identify the two vector spaces.

With this in mind, setting $F : M \rightarrow M \times N$ to be $F(p) = (p, f(p))$, let us analyze what is $F_{*}(v)$ under the identification above. As commented, this would be $(\pi_{1 *}(F_{*}(v)), \pi_{2 *}(F_{*}(v))) = ((\pi_1 \circ F)_{*}(v), (\pi_2 \circ F)_{*}(v)) = (v, f_{*}(v))$ seeing as how $\pi_1 \circ F$ is the identity of $M$ and $\pi_2 \circ F = f$.

The takeaway lesson from this is that in differential geometry there are a lot of natural identifications going on all the time, and this is a frequent source of confusion to beginners. It is good to think thoroughly about them until you are more comfortable with differential geometry.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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