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Oct 25, 2011 at 15:56 comment added Tom Goodwillie Yes, that's right.
Oct 25, 2011 at 13:22 comment added Ritwik One further thing. Requiring $\pi_{\mathcal{D}}$ to be a submersion is a sufficient criteria, but not a necessary one. I simply need $\pi_{\mathcal{D}}$ to be transverse to $X$ (which is guaranteed if it is a submersion).
Oct 25, 2011 at 11:10 comment added Tom Goodwillie Yes, I believe so.
Oct 25, 2011 at 2:48 comment added Ritwik I have edited the question slightly. According to your argument, the answer to my question should now be yes, since $\pi_{D}$ is a submersion. Am I correct?
Oct 22, 2011 at 0:04 comment added Tom Goodwillie I did not express myself very well. I meant that $X$ is parametrized by pairs $q=(x_0,y_0)$, and that for any given point $q$ the manifold $H_q$ fails to be transverse to $X$ (where it meets $X$, namely at the point in $X$ corresponding to $q$). Does that make sense?
Oct 21, 2011 at 19:08 comment added Ritwik Thank you for the answer. But I have a question about the counter example. Here your X also depends on the point q. So as you change q, both the X and the H move. In my question, X does not depend on q. As you change q only H should change. X should remain fixed. So do you still think this is a counter example to my question?
Oct 21, 2011 at 18:59 vote accept Ritwik
Oct 21, 2011 at 15:39 history answered Tom Goodwillie CC BY-SA 3.0