Finding the inverse of the differential $d\pi_{i_p} : T_p( M_1 \times \dots \times M_k) \to T_{p_i}M_i$

Question

Suppose that $M_1, \dots, M_k$ are smooth manifolds. Show that for each $i$ the projection $\pi_i : M_1 \times \dots \times M_k \to M_i$ is a smooth submersion.

I've shown the smoothness of the projection map earlier so to prove this I need to show surjectivity of the differential $d\pi_{i_p}$ at $p=(p_1, \dots, p_k) \in M_1 \times \dots \times M_k$.

My friend suggested me to define a map $i_{M_i}:M_i \to M_1 \times \dots \times M_k$ as $i_{M_i}(x)=(x^1, \dots x, \dots,x^k)$ where each except the i'th position is fixed.

With this map I think I can do the proof as follows. $$(d\pi_{i_p} \circ di_{M_{i_{p_i}}})(v)f = d(\pi_i \circ i_{M_i})_{p_i}(v)f = v(f \circ \pi_i \circ i_{M_i})= v(f)$$ which implies that $di_{M_i}:T_{p_i}M_i \to T_p (M_1, \dots, M_k)$ is an inverse for $d\pi_{i_p}$ and so $d\pi_{i_p}$ must be surjective.

What is confusing to me is this map $i_{M_i}$. What is the purpose of it and why even consider it? It's like the inverse of the projection map or something? I don't think I could have done this without it.

Answer 1

Yes, it is a "right" inverse of $\pi_i$, because $\pi_i\circ i_{M_i}$ is equal to the identity map $Id_{M_i}$ of $M_i$. Knowing that, and applying the chain rule, one gets the same result that you obtained:

$$Id_{T_{p_i}M_i} = d_{p_i}(Id_{M_i}) = d_{p_i}(\pi_i\circ i_{M_i}) = d_p\pi_i \circ d_{p_i}i_{M_i}$$

Which readily shows that $d_p\pi_i$ has a right inverse.

In differential geometry (and other fields), a right inverse of $\pi_i$ is called a "section" of the map $\pi_i$.

A submersion $\pi: M \rightarrow N$ is a map which "projects" the manifold $M$ onto $N$ in such a way that the fiber of each point $q\in N$ is a submanifold of $M$ (kinda like a division of $N$ into "leaves", or a "foliation", though that has its own meaning).

A differentiable section consists on "choosing" for each $q\in N$, a point $s(q)\in M$ in the fiber of $q$ ("above" $q$), that is, such that $\pi(s(q)) = q$. If the section is differentiable, it looks like you are "embedding" $N$ into a "leave" of $M$ or a "cut" of $M$ (that's why it's called a "section", it's like a slice of ham if for example $N$ is a surface).

Answer 2

What you have done is right, although I guess you don’t appreciate the technique.

If $f$ and $g$ are functions such that $f \circ g$ is surjective, then $f$ is surjective. It is asked to show that $f=d\pi_{i_p}$ is surjective and your calculations show that $g= di_{M_{i_{p_i}}}$ is a required map.