I am trying to prove that $$T_{(p,q)}(M\times N)\cong T_pM\oplus T_qN$$

We define: $$\Phi:T_{(p,q)}(M\times N)\to T_pM\oplus T_qN:v\mapsto(d_{(p,q)}\pi_M v,d_{(p,q)}\pi_N v)$$ and $$\Psi:T_pM\oplus T_qN\to T_{(p,q)}(M\times N), (v,w)\mapsto d(\iota_M)_p(v) +d(\iota_N)_q(w),$$ where $\iota_M : M\to M\times N$ sends $M$ to $M \times \{q\}$ and where $\iota_N : N\to M\times N$ sends $N$ to $N\times \{q\}$

I have trouble in showing that $\Psi \circ \Phi=\text{Id}$. Let $f$ be a smooth function on $M\times N$. Compute

\begin{align*} \Psi \circ \Phi(v)f &= \Psi (d_{(p,q)}\pi_M v,d_{(p,q)}\pi_N v)f \\ &= \big(d_{(p,q)}(\iota_M \circ \pi_M) v,d_{(p,q)}(\iota \circ \pi_N) v \big)f \\ &= v(f\circ \iota_M \circ \pi_M )+v(f \circ \iota_N \circ \pi_N) \\ &= v(f\circ \iota_M \circ \pi_M + f \circ \iota_N \circ \pi_N) \end{align*}

But I don't know why the last term is equal to $v(f)$. I looked at many solutions to this problem on stackexchange(this one, for example), but they seem to take this part for granted. Thanks in advance!


It might be easier to see that $\Phi \circ \Psi = \operatorname{Id}$. Then $\Psi$ is injective so you have an isomorphism for dimensional reasons.

To see this take $v \in T_pM, w \in T_qN$, then

\begin{align} \Phi\circ\Psi(v,w) & = \Phi\left(d(\iota_M)_pv + d(\iota_N)_qw\right)\\ & = \left(d(\pi_M)_{(p,q)}\left(d(\iota_M)_pv + d(\iota_N)_qw\right), d(\pi_N)_{(p,q)}\left(d(\iota_M)_pv + d(\iota_N)_qw\right) \right) \\ &= \left(d(\pi_M \circ \iota_M)_pv + d(\pi_M \circ \iota_N)_qw, d(\pi_N\circ\iota_M)_pv + d(\pi_N \circ \iota_N)_qw \right) \end{align}

Now $\pi_M \circ \iota_M$ and $\pi_N \circ \iota_N$ are the identity on $M$ and $N$ respectively so the pushforwards with respect to these maps are also the identity. Since $\pi_M \circ \iota_N$ and $\pi_N \circ \iota_M$ are constant maps, their corresponding pushforwards are zero.


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