Differential of the product smooth map from arbitrary manifold Suppose $f:M\to N\times P$ defined by $f(x)=(f_1(x),f_2(x))$ where $f_1:M\to N$ and $f_2:M\to P$ are smooth maps. Then show that $df=df_1\times df_2$ i.e. $df(X)=(df_1(X), df_2(X))$. ($df$ is the differential of $f$ & $M,N,P$ are smooth manifolds)
I was trying by the formula $df(X)(g)=X(g\circ f)$ where $X$ is a tangent vector and $g$ is a real valued function. But did not able to complete the proof. Please help to do this.
 A: Suppose $f : M \to P \times Q$ is a smooth map defined as $f(x) = (g(x),h(x))$ where $g \in C^{\infty}(M, P)$ and $h \in C^{\infty}(M,Q)$. Let $x \in M$ and $g(x)=p \in P$, $h(x)=q\in Q$. First, we know that $T_{(p,q)}(P \times Q) \cong T_pP \times T_qQ$ by isomorphism
$$
\alpha : T_{(p,q)}(P \times Q) \longrightarrow T_pP \times T_qQ, \quad \alpha(v) = \Big(d(\pi_1)_p(v), d(\pi_2)_q \Big),
$$
where $\pi_1 : P \times Q \to P$ and $\pi_2 : P \times Q \to Q$ are the canonical projections.
Let's define $dg_x \times dh_x : T_xM \to T_pP \times T_qQ $ as the map $(dg_x \times dh_x)(v) = \Big(dg_x(v),dh_x(v)\Big)$. Since the codomain of $df_x : T_xM \to T_{(p,q)}(P \times Q)$ and $dg_x \times dh_x$ are not really the same, what we really need to show is that the composition map
$$
T_xM \xrightarrow{df_x} T_{(p,q)}(P \times Q) \xrightarrow{\alpha} T_pP \times T_qQ
$$
is equal to $dg_x \times dh_x$. But this is just a modest computation using the chain rule :
\begin{align}
\alpha \circ df_x (v) &=  \big( d(\pi_1)_p \circ df_x(v), d(\pi_2)_q \circ df_x(v) \big)\\
&= \big( d(\pi_1 \circ f)_x(v),d(\pi_2 \circ f)_x(v) \big) \\
&= \big( dg_x(v),dh_x(v) \big)\\
&= dg_x \times dh_x(v).  
\end{align}
See here for a more general case.
