Inverse of a map $T_{(p,q)}(X \times Y) \to T_p X \times T_p Y$

Question

I'm trying to show that the maps defined in https://math.stackexchange.com/a/413846/500094 are mutual inverses. I have problem with one direction:

$(g\circ f)(v)=g(d(\pi_X)_{(p,q)}v,d(\pi_Y)_{(p,q)}v)=d(\iota_X)_p(d(\pi_X)_{(p,q)}v)+d(\iota_Y)_q(d(\pi_X)_{(p,q)}v)=\\d(\iota_X\circ\pi_X)_{(p,q)}v+d(\iota_Y\circ\pi_Y)_{(p,q)}v=v+v=2v\ne v$

At which point am I mistaken?

Answer 1

In general, if $W$ is a f.d. vector space and $P_V\colon W \to W$ is a projection map with image $V\subset W$ and kernel $U$, then $W \cong U \oplus V$ and the map $P_U+P_V = \text{id}_W$ (where $P_U\colon W \to W$ is the projection map with image $U$).

In this case, take $L = d(\iota_X\circ \pi_X)_{(p,q)}$.

1. Show $L$ is a projection map ($L\circ L = L$).

2. What is the image of $L$?

3. What is the kernel of $L$?

4. What map appearing in your question above is a formula for projection onto the kernel of $L$?

If you can answer these questions, you will have a nice answer to the problem.

Update: here are some extra details:

1. We can use properties of pushforward and the fact that $\pi_X\circ \iota_X = \text{id}_X$ to show that $$ L\circ L = d(\iota_X\circ \pi_X)_{(p,q)}\circ d(\iota_X\circ \pi_X)_{(p,q)} = d(\iota_X\circ (\pi_X\circ \iota_X)\circ \pi_X)_{(p,q)} = d(\iota_X\circ \pi_X)_{(p,q)} = L$$ so $L$ is a projection map.

2. The image of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the subspace $V = T_{(p,q)}X\times \{q\}$, the tangent space at $(p,q)$ of the submanifold $X\times \{q\}$. In short, the image of this map is the set of tangent vectors at $(p,q)$ to paths of the form: $$ \iota_X\circ \pi_x\circ \gamma: \mathbb{R} \to X\times Y$$ where $\gamma$ is a path in $X\times Y$ with $\gamma(0) = (p,q)$. Such a $\gamma$ can be written as $\gamma(t) = (\gamma_1(t),\gamma_2(t))$ for $\gamma_1\colon \mathbb{R}\to X$ and $\gamma_2\colon \mathbb{R}\to Y$. You should verify that $$\iota_X\circ \pi_x\circ \gamma(t) = (\gamma_1(t),q)$$ Thus the image of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the set of vectors in $T_{(p,q)}X\times Y$ that can be written in the form $$d(\iota_X\circ \pi_X)_{(p,q)}\gamma'(0) = \frac{d}{dt}|_{t=0}(\gamma_1(t),q) = (\gamma_1'(0),0)$$ which is the desired tangent space.

3. The kernel of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the subspace $U = T_{(p,q)}\{p\}\times Y$. I leave the justification to you.

4. So which map above gives projection onto $U$? It's $d(\iota_Y\circ \pi_Y)_{(p,q)}$. The proof is similar to items 1. and 2. above. In particular, this map is NOT the identity map.