Let $M$ be a manifold equipped with a pair of surjective submersions $N_1 \stackrel{p_1}{\leftarrow} M \stackrel{p_2}{\rightarrow} N_2$ where $dim N_1 = dim N_2 = n$. Then we can find, for any point $m\in M$, a chart $U_1$ around $p_1(m)$ and a local section $s\colon U_1 \to M$ of $p_1$ such that $s(p_1(m)) = m$ and $s(U_1)$ is transverse to the fibres of $p_2$. In thinking about this we can clearly reduce to the case $N_1 = N_2 = \mathbb{R}^n$. We can even restrict attention to $M = \mathbb{R}^m$, and then it becomes a problem of linear algebra, namely finding a basis on $\mathbb{R}^m$ for which the submersions are *both* projections onto $n$ coordinates (we already know this separately).

I suspect that for infinite-dimensional vector spaces, and Frechet spaces in particular (really anything above Hilbert spaces in the usual hierarchy) this sort of result will not hold, and so for Frechet manifolds one cannot construct the analogous local section.

In more detail, I'm fairly sure that given a diagram of Frechet spaces

$$V_1 \stackrel{pr_1}{\leftarrow} V_1 \times F_1 \simeq W_1 \simeq V_2 \times F_2 \stackrel{pr_1}{\to} V_2$$

where $V_1$ is known to be isomorphic to $V_2$, one cannot in general find a (nonlinear?) map $F\colon V_1 \to V_2\times F_2$ (which is a section, and passes through the origin) such that $pr_1\circ F\colon V_1 \to V_2$ is ~~injective~~ *an isomorphism*. However, I'd like to see a counterexample (or, if I'm wrong, a proof that we can do it!). (EDIT: the map might need be be non-linear in this case because the definition of submersions doesn't use tangent spaces like in the finite-dimensional case. But linear would be good.)

VERSION 2: well it turns out that I have, in addition to the $V_i$s being isomorphic, I have $F_1$ isomorphic to $F_2$. Kudos to Andrew for guessing this would be the case for the application I have in mind.