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Let $P,Q,M$ be smooth manifolds, $f:P \rightarrow M$, $g:Q \rightarrow M$ be two smooth maps, which are transverse to each other. Denote by $Z$ the fiber product of $f$ and $g$, $Z=\{(p,q) \in P \times Q \; | \; f(p)=g(q)\}$. Since $f$ and $g$ are transverse, this is an embedded submanifold of $P \times Q$ of dimension $\dim P+\dim Q - \dim M$.

Let $(p,q) \in Z$ and $x:=f(p)=g(q)$. Consider the map $$ \psi: T_{p}P \times T_{q}Q \longrightarrow T_{x}M $$

given by $(u,v) \mapsto (d_{p}f)(u)-(d_{q}g)(v)$. The claim is that its kernel, $\ker(\psi)$ is given by

$$ T_{(p,q)}Z $$

where, as usual, we identify $T_{(p,q)}Z$ as a sub-vector space of $T_{p}P \times T_{q}Q$.

I was able to show that $T_{(p,q)}Z$ is contained in $\ker(\psi)$, but the reverse inclusion I could not prove. Can anyone help me out?

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1 Answer 1

up vote 0 down vote accepted

It is almost embarrassingly simple...

Since $f$ and $g$ are transverse the map $\psi$ above is surjective. By linear algebra we have

$$ \dim (T_{p}P \times T_{q}Q)=\dim (\ker(\psi))+\dim M $$

and hence $\dim (\ker(\psi))=\dim P +\dim Q -\dim M$.

As $T_{(p,q)}Z \subseteq \ker(\psi)$ and $\dim T_{(p,q)}Z=\dim (\ker(\psi))$ we must have $\ker(\psi)=T_{(p,q)}Z$.

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