Local bisections of submersions Consider two surjective submersions $s, t : G \to M$.  The following proposition appears in many places:

For any $g$ in $G$ there is an open $U \subseteq M$ and a smooth section $\sigma : U \to G$ of $s$ such that $\sigma(s(g)) = g$ and $t \cdot \sigma$ is a diffeomorphism onto its image.  

I am trying to understand the proof of this claim.  Every proof I can find is essentially the same; here's the one from Moerdijk and Mrčun's Introduction to Foliations and Lie Groupoids (Prop. 5.3):

Choose a subspace $V$ of $T_{g}G$ which is complementary to both $\ker(ds)_{g}$ and $\ker(dt)_{g}$.  Choose a local section $\sigma : U \to G$ of $s$ such that $(d\sigma)_{s(g)}(T_{s(g)}M) = V$.  It follows that $d(t \cdot \sigma)_{s(g)}$ is an isomorphism, so we can shrink $U$ if necessary so that $t \cdot \sigma$ becomes an open embedding.

I have two questions.


*

*Why can we choose a single subspace $V$ that's complementary to both $\ker(ds)_{g}$ and $\ker(dt)_{g}$?  Since $s$ and $t$ are submersions we have
$$
T_{g}G 
\cong \ker(ds)_{g} \oplus T_{s(g)}M
\cong \ker(dt)_{g} \oplus T_{s(g)}M
$$
But why can we choose these isomorphisms such that they send $T_{s(g)}M$ to the same subspace of $T_{g}G$?

*Why can we choose the image of $(d\sigma)_{s(g)} : T_{s(g)}M \to T_{g}G$ to be $V$?  If we were just looking for a section of $s$ such that $\sigma(s(g)) = g$ then we'd choose some coordinates such that $s$ takes the form
$$
(x^{1}, \dotsc, x^{k}, x^{k+1}, \dotsc, x^{n})
\mapsto
(x^{1}, \dotsc, x^{k})
$$
and then construct $\sigma$ using the section
$$
(x^{1}, \dotsc, x^{k})
\mapsto
(x^{1}, \dotsc, x^{k}, 0, \dotsc, 0)
$$
Probably the way to make $V$ the image of $(d\sigma)_{s(g)}$ is to use something other than the zero section here, or to choose coordinates adapted to $V$ somehow.
 A: *

*Suppose that $W\subset {\mathbb R}^n$ is a $k$-dimensional subspace. Then for an open and dense subset $U_W$ in the Grassmannian $Gr_{n-k}({\mathbb R}^n)$, every $V\in U_W$ is complementary to $W$. (This is a pleasant linear algebra exercise proven, for instance, by induction on dimension or, if you want to use a big hammer, -- by Sard's theorem. This exercise is a manifestation of the "generic transversality" in differential topology.) Thus, given two (or even countably many) $k$-dimensional subspaces $W_1, W_2\subset  {\mathbb R}^n$, $U_{W_1}\cap U_{W_2}$ is nonempty, hence, there exists $V\in U_{W_1}\cap U_{W_2}$, which is then complementary to both $W_1, W_2$. 

*Yes, you have to use a different section. Since $s$ is linear (in your local coordinates), this is again a linear algebra exercise:
Let $A: {\mathbb R}^n\to {\mathbb R}^{n-k}$ be a surjective linear map with the kernel $K$. Then for every subspace $V\subset {\mathbb R}^n$ complementary to $K$, there exists a linear map $B: {\mathbb R}^{n-k}\to V$ such that $A\circ B=Id$. (Hint: Verify that $A$ restricted to $V$ is both injective and surjective.) 
Use the map $B$ as the section $\sigma$. 
