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While reading the book Introduction to Smooth Manifolds by John M. Lee I get confused by the statement of the Parametric Transversality Theorem (see last paragraph). Let $F:\mathbb R\times \mathbb R\to \mathbb R^2$ be defined by $F(x,s) = (0,s)$. Then $F$ is smooth, which I believe is transverse to the submanifold $X =\mathbb R\times \{0\}$. However, for the smooth family of maps $\{F_s\}$ defined by $F_s(x) = F(x,s)$, there is no $s\in \mathbb R$ such that $F_s$ is transverse to the embedded submanifold $X$. This contradicts the Parametric Transversality Theorem. So there must be something that I missed. Any help is appreciated.

I give below a relevant definition and a theorem in the book to standardize the discussion.

$F$ transverse to $S$. If $F:N\to M$ is a smooth map and $S\subseteq M$ is an embedded submanifold, we say $F$ is transverse to $S$ if for every $x\in F^{-1}(S)$, the spaces $T_{F(x)}S$ and $dF_x(T_xN)$ together span $T_{F(x)}M$.

Parametric Transversality Theorem. 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 $s\in S$, the map $F_s:N\to M$ is transverse to $X$.

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Au contraire. It's transverse (vacuously) for all $s\ne 0$.

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