Manifold Transversality

Lee Smooth Manifolds

Global Characterization of Graphs (Lee 6.32): Suppose $M$ and $N$ are smooth manifolds and $S\subseteq M\times N$ is an immersed submanifold. Let ${\pi }_M$ and ${\pi }_N$ denote the projections from $M\times N$ onto $M$ and $N$, respectively. The following are equivalent: (a) $S$ is the graph of a smooth map $f:M\to N$. (b) ${\pi }_M{|}_S$ is a diffeomorphism from $S$ onto $M$. (c) For each $p\in M$, the submanifolds $S$ and $\{p\}\times N$ intersect transversely in exactly one point. If these conditions hold, then $S$ is the graph of the map $f:M\to N$ defined by $f={\pi }_N\circ ({\pi }_M{|}_S{)}^{-1}$.

Proof) $(a)\Rightarrow (b)$ is essentially true by how the smooth structure of graphs is defined. $(b)\Rightarrow (a)$ is true as if ${\pi }_M{|}_S$ is a diffeomorphism, $f:={\pi }_N\circ ({\pi }_M{|}_S{)}^{-1}$ is a well-defined smooth map from $M$ to $N$such that $S$ is, as a set, the graph of $f$. By the uniqueness of the submanifold structure of embedded submanifolds, $S$ must be precisely the graph of $f$. $(b)\Rightarrow (c)$ holds as, for any $p\in M$, $d({\pi }_M{|}_S{)}_{(p,f(p))}:T_{(p,f(p))}S\to T_{p}M$ is an isomorphism, and as $d({\pi }_M{)}_{(p,f(p))}$ is simply a projection, this implies $T_{(p,f(p))}S+T_{(p,f(p))}N=T_{(p,f(p))}(M\times N)$. It suffices to show $(c)\Rightarrow (b)$. Reversing our previous logic, we get that $d({\pi }_M{|}_S{)}_{(p,f(p))}$ is surjective and, by dimensionality reasons, must be an isomorphism. Thus, ${\pi }_M{|}_S$ is a local diffeomorphism from $S$ onto $M$, and must be injective as $S$ and $\{p\}\times N$ always intersect in precisely one point. This forces ${\pi }_M{|}_S$ to be a diffeomorphism, completing our proof.

Parametric Transversality Theorem (Lee 6.35): Suppose $N$ and $M$ are smooth manifolds, $X\subseteq M$ is an embedded submanifold, and $\{F_s:s\in S\}$ is a smooth family of maps from $N$ to $M$. If the map $F:N\times S\to M$ is transverse to $X$, then for almost every $s\in S$, $F_s:N\to M$ is transverse to $X$.

Proof) As $F$ is transverse to $X$, $W=F^{-1}(X)$ is an embedded submanifold of $N\times S$. Let $\pi :W\to S$ be the projection onto $S$. It suffices to show that if $s\in S$ is a regular value of $\pi$, $F_s$ is transverse to $X$.

Let $p\in F_s^{-1}(X)$ be arbitrary and $q=F_s(p)$. We aim to show that

$$T_{q}M=T_{q}X+d(F_s{)}_p(T_{p}N)$$

By hypothesis, we have that

$$d{\pi }_{(p,s)}(T_{(p,s)}W)=T_{s}S$$

Furthermore, as $F$ is transverse to $X$, we have that

$$T_{q}M=T_{q}X+d{F}_{(p,s)}(T_{(p,s)}(N\times S))$$

Let $k$ be the codimension of $X$ and $\Phi :U\to {\mathbb{R}}^k$ be a local defining function of $X$ where $U$ is a neighborhood of $q$. Then, $T_{q}X=\text{Ker}(d{\Phi }_q)$ and $T_{(p,s)}W=\text{Ker}(d{\Phi }_q\circ d{F}_{(p,s)})$, and thus

$$d{F}_{(p,s)}(T_{(p,s)}W)\subset T_{q}X$$

We can now complete the proof. It suffices to show that

$$d{F}_{(p,s)}(T_{(p,s)}(N\times S))\subset T_{q}X+d(F_s{)}_p(T_{p}N)$$

Let $(v,w)$ be an arbitrary element of $T_{(p,s)}(N\times S)$. As $d{\pi }_{(p,s)}(T_{(p,s)}W)=T_{s}S$, there must be some $v^{\prime }\in T_{p}N$ such that $(v^{\prime },w)\in T_{(p,s)}W$. By linearity,

$$d{F}_{(p,s)}(v,w)=d{F}_{(p,s)}(v^{\prime },w)+d{F}_{(p,s)}(v-v^{\prime },0)$$

Here, $d{F}_{(p,s)}(v^{\prime },w)\in d{F}_{(p,s)}(T_{(p,s)}W)\subset T_{q}X$, and if we let $i:N\to N\times S$ be the inclusion $n\mapsto n\times \{s\}$, $(v-v^{\prime },0)=d{i}_p(v-v^{\prime })$, and as $F_s=F\circ i$,

$$d{F}_{(p,s)}(v-v^{\prime },0)=d{F}_{(p,s)}\circ d{i}_p(v-v^{\prime })=d(F_s{)}_p(v-v^{\prime })\in d(F_s{)}_p(T_{p}N)$$

we are done.