# showing pushfarward

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?

• 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. – Pedro Apr 19 '15 at 10:55
• Sorry, but how does it help me? (I proved F is smooth, but I can not prove the second part) – 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 $\pi_2 \circ F = f$.