Differential of the map $(f_1, f_2) : M \to N_1 \times N_2$.

Question

Let $M, N_1$ and $N_2$ be a smooth manifolds and define the smooth maps $f_1 :M \to N_1, f_2:M \to N_2$. Let $p \in M$ and define the differential $d(f_1,f_2)_p : T_pM \to T_{(f_1(p), f_2(p))}(N_1 \times N_2)$. Show that $d(f_1,f_2)_p(v)=df_{1_p}(v)+df_{1_p}(v)$ for any $v \in T_pM$.

My approach was to show that $T_pM\xrightarrow{d(f_1,f_2)_p} T_{(f_1(p), f_2(p))}(N_1 \times N_2) \xrightarrow{\alpha}T_{f_1(p)}N_1 \oplus T_{f_2(p)}N_2$ commutes, where $\alpha$ is the isomorphism between $T_{(f_1(p), f_2(p))}(N_1 \times N_2)$ and $T_{f_1(p)}N_1 \oplus T_{f_2(p)}N_2$.

What I got is that $$\alpha \circ d(f_1,f_2)_p(v)=(d\pi_{1_{f_1(p)}} \circ d(f_1,f_2)_p(v), d\pi_{2_{f_2(p)}} \circ d(f_1,f_2)_p(v)) = (d(\pi_1 \circ (f_1,f_2)_p(v))_p, d(\pi_2 \circ (f_1,f_2)_p(v)))$$ which should equal $(df_{1_p}(v), df_{2_p}(v))$ and by the isomorphism identification $(df_{1_p}(v), df_{2_p}(v)) = df_{1_p}(v) + df_{2_p}(v)$?

Answer 1

Since $\pi_i \circ (f_1, f_2) = f_i$ for $i=1, 2$, the chain rule implies

\begin{align} d(\pi_i)_{(f_1(p), f_2(q))} \circ d (f_1, f_2)_{(p, q)} &= d (\pi_i \circ (f_1, f_2))_{(p,q)} = d(f_i)_p. \end{align}

Hence $$ \alpha \circ d(f_1,f_2)_p(v)= (d(f_1)_p (v), d(f_2)_p (v))$$

which is what you want.