# Projection $\pi_i : M_1 \times \cdots \times M_n \to M_i$ is a Smooth Submersion?

I want to show the following.

Let $\{M_i|i\in I\}$ be a set of smooth manifolds and $p:=(p_1,\cdots,p_n)\in M:= M_1\times\cdots\times M_n$ be a point in the smooth product manifold and

$$\pi_i:M\rightarrow M_i,\quad p\mapsto p_i$$

be the projection onto the $i$-th factor. Then $\pi_i$ is a smooth submersion.

I want to show this proposition without choosing coordinate charts but merely by showing that the differential $d\pi_i|_p:T_pM\rightarrow T_{p_i}M_i$ is surjective.

So I pick a $w_p\in T_{p_i}M_i$ and define a $v_p\in T_pM$ via $v_pf:=w_pg$, where $f\in\mathcal{C}^\infty(M),g\in\mathcal{C}^\infty(M_i)$, such that $f|_{M_i}=g$. Since for two $g,g^\prime\in\mathcal{C}^\infty(M_i)$ with $g|_U=g^\prime|_U$ on some small neighbourhood $U\subset M_i$ of $p_i$ we have $w_pg=w_pg^\prime$, the tangent vector $v_p$ is well-defined.

Then we have

$$d\pi_i|_p(v)f=v_p(f\circ\pi_i)=v_pf|_{M_i}=w_pg.$$

Is this already enough to show the surjectivity?

• You can alternatively consider the map $\alpha : M_i \to M, \alpha(x) = (a_0, ... ,a_{i-1},x,a_{i+1}, ... ,a_n)$ with fixed $a_j \in M_j$. You have $\pi_i \circ \alpha = id$. – Paul Frost Jun 1 '18 at 12:32

Your did pick the right vector : $v_pf:=w_{p_i}(f|_{M_i})$. But here you should be more specific about what the function $f|_{M_i}= f \circ \iota_i$ mean. Because there are uncountably many copies of $M_i$ inside $M$, that is we can have many inclusion $\iota_i : M_i \hookrightarrow M$ as $\iota_i(x) = (c_1,\cdots,c_{i-1},x,c_{i+1},\cdots,c_n)$. But we will see that the correct inclusion is $\iota_i(x) = (p_1,\cdots,p_{i-1},x,p_{i+1},\cdots,p_n)$. The other choice of injection is fail because the operator $v_p : C^{\infty}(M) \to \mathbb{R}$ correspond to it, is not a derivation (it fails to satisfy Leibniz Rule). I'll put the proof in the note below.
Now, for any $g \in C^{\infty}(M_i)$ and $w_{p_i}\in T_{p_i}M_i$, the vector $v_p \in T_pM$ defined as $v_p(f):=w_{p_i}(f \circ \iota_i)$, with $\iota_i(x) = (p_1,\cdots,p_{i-1},x,p_{i+1},\cdots,p_n)$, satisfy $$d \pi_i (v_p)g =v_p (g \circ \pi_i) = w_{p_i} (g \circ \pi_i)|_{M_i} = w_{p_i} (g \circ \pi_i \circ \iota_i) = w_{p_i} g.$$
$\textbf{Note : }$
$\textbf{The vectors } v_{p}f:=w_{p_i}(f \circ \iota_i), \textbf{ where } \iota_i (x):= (c_1,\dots,c_{i-1},x,c_{i+1},\dots,c_n) \textbf{ and } \iota_i(p_i)\neq p = (p_1,\dots,p_n), \textbf{ is not a derivation}$.
Let $f,g \in C^{\infty}(M)$ be arbitrary. By def we have \begin{align*} v_p(fg) &= w_{p_i}(fg)|_{M_i} = w_{p_i} ((fg) \circ \iota_i) = w_{p_i} (f|_{M_i}g|_{M_i})\\ &= f(c_1,\dots,c_{i-1},p_i,c_{i+1},\dots,c_n) v_pg + g(c_1,\dots,c_{i-1},p_i,c_{i+1},\dots,c_n) v_pf \end{align*} which is not necessarily equal to $f(p)v_pg + g(p) v_pf$.
As @PaulFrost says, we can alternatively prove the surjectivity of $d\pi_{i}$ by find the right-inverse for it. That is a linear map $L : T_{p_i}M_i \to T_pM$ such that $d\pi_i \circ L = \text{Id}_{T_{p_i}M_i}$. The appropriate right-inverse is $$d\iota_i : T_{p_i}M_i \to T_pM,$$ since for any $w_{p_i} \in T_{p_i}M_i, f\in C^{\infty}(M_i)$, $$d\pi_i \circ d\iota_i (w_{p_i}) f = d(\pi_i \circ \iota_i)(w_{p_i})f = w_{p_i}(f \circ \pi_i \circ \iota_i) = w_{p_i}f.$$ However, the similar construction is used to prove that actually $T_pM = T_p(M_1 \times \cdots \times M_n) \cong T_{p_1}M_1 \times \cdots \times T_{p_n}M_n$. (See here for example)