# Prove closed of dimension one of $X\times I$.

Suppose $F(x,t): X\times I \rightarrow R$ is a homotopy of Morse functions. That is, $f_t: X \rightarrow R$ is Morse for every $t$. Show that the set $C = \{(x,t)\in X\times I : d(f_t)_x = 0\}$ forms a closed, smooth submanifold of dimension one of $X\times I$. Assume the homotopy is constant near the ends of I and use an open interval.

What I have done so far is that showed it is a smooth manifold by inverse function theorem. But I don't know how to show it is closed and has dimension 1.

Thank you very much for your guidance!

• "closed, smooth submanifold of dimension one of $X \times I$ could also be phrased "one-dimensional, closed, smooth submanifold of $X \times I$," if that helps. – Daniel McLaury May 31 '13 at 4:38
• Hi @DanielMcLaury - Thank you very much for your help! But would you mind sharing me some insight with how to prove it is closed and with dimension 1? Thanks! – 1LiterTears May 31 '13 at 5:13
• @DanielMcLaury Got it! Thank yoU! – 1LiterTears May 31 '13 at 6:11

Submanifold: When $dF=0, d^2 F \neq 0$, we know the Jacobian of $dF$ is not singular. Hence by inverse function theorem, it is a submanifold.
Closed: The given condition {$(x,t) \in X \times I:d(f_t)_x=0$} is closed, since all the limit points satisfy the condition and therefore is included in the set.
1-dimension: By Morse function condition, because $d^2 F \neq 0$, that means given $df_x = 0$, that means $df_{x^\prime} \neq 0$ for all $x^\prime$ in any neighborhood of $x$. Therefore, the set can be only varied by $t$, which suggests it is 1-dimensional.