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It is a problem from Guillemin and Pollack.

Suppose that $X$ is a submanifold of $\mathbb R^N$. Show that almost every vector space $V$ of any fixed dimension $l$ in $\mathbb R^N$ intersects $X$ transversally. [HINT: The set $S\subset (\mathbb R^N)^l$ consisting of all linearly independent $l$-tuples if vectors in $\mathbb R^N$ is open in $\mathbb R^{Nl}$, and the map $\mathbb R^l\times S\rightarrow \mathbb R^N$ defined by $[(t_1,\dots,t_l),v_1,\dots,v_l]\mapsto t_1v_1+\dots t_lv_l$ is a submersion.]

Suppose that $F:X\times S\to Y$ is a smooth map of manifolds and $Z$ is a submanifold of $Y$, all manifolds without boundary. If $F$ is transverse to $Z$ then for almost every $s\in S$ the map $f_s : x\mapsto F(x,s)$ is transverse to $Z$.

I have shown that the function in the hint is indeed a submersion and by theorem above,for every element the map is transversal and it is affine subspace of $\mathbb R^N$ Is it enough to prove it? I am not sure.

Jale'de jaled
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    Is $V$ a linear (affine) subspace of $\Bbb R^N$? – Berci Jan 29 '21 at 18:40
  • yes................ – Jale'de jaled Jan 29 '21 at 18:41
  • The map is not submersion at $t=0$, there the image of the derivative is the span of the $v_i$'s. Besides $V$ and $T_xX$, $x\in V\cap X$, must generate $\mathbb R^N$, which is imposible if dimensions do not fit. Suppose $x=0\in X$, a curve in $\mathbb R^3$. Then a line $V$ through the origin can never be transversal to $X$ with this definition. The typical resource to moving the line to make it disjoint of $X$ gives an affine line! If one alows affine $V$, the argument works: just translate $X$ off the origin. – Jesus RS Jan 13 '23 at 11:37

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