# Preimage Lemma for transverse map. Help with some passages

I'm on my way proving the Preimage Lemma for a transverse smooth map but I've encountered some problems with two passages:

Let $f\colon M \to N$ be a smooth map transverse to a submanifold $L$ of $N$ of codimension $k$ and $f^{-1}(L)$ is not empty. Then $f^{-1}(L)$ is a codimension $k$ submanifold of $M$

The proof given during lecture is the following:

Let $f(p)=q$, $q \in L$. For some neighbourhood $V$ of $q$, $L\cap V \simeq \mathbb{R}^{n-k}\cap V'$ where we see $\mathbb{R}^{n-k}$ as $(x_1,\dots,x_{n-k},0,\dots,0)$ in $\mathbb{R}^n$). Let $\pi\colon \mathbb{R}^n\to \mathbb{R}^k$ be the projection to the last $k$-coordinates. By transversality we have that $0 \in \mathbb{R}^k$ is a regular value of the composition $$U \xrightarrow{f} V \simeq V' \xrightarrow{\pi} \mathbb{R}^k$$ Hence for an open neighbourhood $U$ in $M$, $f^{−1}(L)\cap U$ is a codimension $k$ submanifold of $U$ (by the Regular Value Theorem). It follows that $f^{−1}(L)$ is a codimension $k$ submanifold of $M$.

my doubts are:

1) why $0$ is a regular value. My opinion is that the preimage of $0$ via the projection and the diffeomorphism in the middle is (in the non trivial case) a point of $f(M)\cap L$, but I don't see why by transversality, the rank of $f$ in a preimage of such point should be maximal.

2) where does the $U$ come from and how the Regular Value Thm is applied. In my opinion I'd conclude that the preimage of such $0$ is a submanifold, but I don't know how to prove the more general assertion in the proof (why in the proof we go from $f^{-1}$ of a point to $f^{-1}(L)$?).

thanks in advance for any help!

1) The rank of $\pi\circ f$, not of $f$. Look at differentials. We may identify $\pi$ with its differential. The point is in showing that $\ker \pi\cap \mathrm{im}(d_pf)=0$. Once you have this you just apply the usual dimension formulas for kernel and image of a linear operator. Remember that you can use $\mathrm{rank} (\pi \circ d_pf)\le \mathrm{min}(\mathrm{rank}\pi,\mathrm{rank}d_p f)=k$.
2) Again, take care you are taking about $\pi\circ f$. Then the point is that $(\pi\circ f)^{-1}(0)=f^{-1}(L)$ (at least locally).
• Hi, sorry for late commenting this answer, but I was quite busy trying figure out your reasoning. I don't see how to prove that $\ker \pi \cap Im(d_pf) = 0$ could you give me some more hints? – Luigi M Aug 22 '15 at 16:51
• When you say that $L\cap V\simeq \mathbb R^{n-k} \cap V^\prime$ you get also, looking at tangent spaces, that $T_qL\simeq \mathbb R^{n-k}$, with ${\mathbb R}^{n-k}$ given by the first coordinates. This coincides precisely with the kernel of $\pi$, since $\pi$ projects on the last $k$ coordinates. On the other hand transversality implies $\mathrm{im}d_p f\oplus T_qL=\mathbb R^k$, and I think this is it. – N. Ciccoli Aug 23 '15 at 8:49
• Nope, transversality means $\mathbb{R}^n = Im d_pf + T_qL$ i.e no direct sum. Anyway I managed to work out what I needed thanks to your hints :) – Luigi M Aug 24 '15 at 14:46