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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!

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1 Answers1

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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.

Charlie Cifarelli
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