Tangent space to a product Can you explain this question explicitly. This is a little bit difficult for me, but I want to learn how to solve. Thank you for help. 

If $M$ and $N$ are manifolds, let $\pi_1:M\times N\to M$ and $\pi_2: M\times N\to N$ be the two projections.  Prove that for $(p,q)\in M\times N$, $$(\pi_{1*},\pi_{2*}):T_{(p,q)}(M\times N)\to T_pM\times T_qN$$ is an isomorphism.

 A: Hint: You have the obvious maps $i_1:M\hookrightarrow M\times N$ and $i_2:N\hookrightarrow M\times N$. Let $i_\ast=(i_1)_\ast\oplus (i_2)_\ast$. Show using the functorality and linearity of the pushforward that $i^\ast$ is a linear inverse to your map.
Further hint: You want to show that $\pi_\ast\circ i_\ast=\text{id}$, where $\pi_\ast=((\pi_1)_\ast,(\pi_2)_\ast)$. To see this, we merely make the following computation:
$$\begin{aligned}\pi_\ast\circ i_\ast &= ((\pi_1)_\ast(i_{1,\ast}+i_{2,\ast}),(\pi_2)_\ast(i_{1,\ast}+i_{2,\ast}))\\ &= (\pi_1(i_{1,\ast})+\pi_1(i_{2,\ast}),\pi_2(i_{1,\ast})+\pi_2(i_{2,\ast}))\\ &= ((\pi_1\circ i_1)_\ast+(\pi_1\circ i_1)_\ast,(\pi_2\circ i_1)_\ast+(\pi_2\circ i_2)_\ast)\\ &= ((\text{id}_M)_\ast+(c)_\ast,(c)_\ast+(\text{id}_N)_\ast)\\ &= (\text{id}_{T_p M},\text{id}_{T_qN})\\ &=\text{id}_{T_p M\times T_q N}\end{aligned}$$
where $c$ is a constant (not the same one in each coordinate). We made use of the chain rule, the additivity of pushforwards, the fact that constants have zero pushforwards etc.
You are technically done since we're in finite dimensions, and any one-sided inverse is a two-sided inverse, but I urge you to do the computation in the other direction to test yourself on understanding.
