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The following text confuses me very much:

(1) For fixed $x \in X$, is my understanding correct?

  • $F_{(x,s)}$ takes $(x,s)$ and gives $f(x)+s$, so $dF_{(x,s)}$ takes $(x,s)$ and gives $df(x)+ds$. But $x$ is fixed, so $df(x) = 0$. Also, $ds$ is just identity. Hence, $\forall v \in T_{(x,s)} (X,S), \exists u \in (X,S): dF_{(x,s)}(u)=v.$ Namely, $u = (x,s).$ Therefore, $dF$ is surjective hence a submersion.

(2) Why $F$ is a submersion of $X \times S$?

(3) And why therefore transversal to any submanifold $Z$ of $\mathbb{R}^M$?

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Thank you~

1LiterTears
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1 Answers1

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Your argument for (1) looks good, except that $u$ should be $(0,v)$, where the first component is the zero tangent vector at $x$, not $x$ itself (since the arguments of $dF$ need to be tangent vectors, tangent to the submanifold $\{x\}\times S$).

For (2), just observe that the tangent space to $X\times S$ at $(x,s)$ includes, as a subspace, the tangent space to $\{x\}\times S$. Since the latter maps, via $dF$, onto the whole tangent space to $\mathbb R^M$, so does the former.

For (3), recall that the desired transversality of $F$ to $Z$ means only that the image of $dF$ and the tangent space to $Z$ together generate the whole tangent space to $\mathbb R^M$. In the case at hand, the image of $dF$ is already this whole tangent space, so we don't even need the tangent space of $Z$ to generate that whole tangent space. (In other words, submersions are transversal to everything.)

Andreas Blass
  • 71,833
  • Thank you so much Andreas. That is very inspiring and very helpful. It clears out many of the confusions I had. Thanks again!! – 1LiterTears Jul 15 '13 at 22:38