Milnor's 'Lectures on h-cobordism theorem' Theorem 7.6 ("Basis Theorem") reads

Suppose $(W;V,V')$ is a triad of dimension $n$ (ie. a cobordism $W$ between $V$ and $V'$) possessing a Morse function $f$ with all critical points of index $\lambda$ and on the same level, and let $\xi$ be a gradient-like vector field for $f$ (ie. $\xi(f)>0$ and $\xi$ matches $\nabla f$ in a coordinate nbd of the critical points of $f$). Assume that $2\leq \lambda \leq n-2$ and that W is connected. ...

(conclusion of the theorem is omitted since my problem does not concern it)

Milnor starts the proof (after some motivation) by raising $f$, by a constant, locally around a critical point $p_1$ to form a new Morse function $f_1$. Then we would have $f_1(p_1)>f(p_1)$. Pick $t_0$ such that $f_1(p_1)>t_0>f(p_1)$ and let $V_0=f^{-1}(t_0)$. Later he claims $V_0$ is connected, "since $W$ is connected".

I am at a lost as to why this reasoning holds. It is certainly not true that a level set of a Morse function on a connected manifold must be connected. I am not sure whether the hypothesis $2\leq \lambda \leq n-2$ can make this true though since frankly I can only visualize the case for 2-manifolds. I am only able to figure out that for one critical point of index between $2$ and $n-2$, the level sets are connected and that this fails for index $1$ and $n-1$.

Any help would be appreciated!


1 Answer 1


The attaching region of an handle of index $n-2 \geq \lambda \geq 2$ is connected so the number of connected components of the (sub)level set of a Morse function can't increase/decrease while your are passing a critical point of such a index. On the other hand:

1) each time you pass a index $\lambda=0$ critical point you $\textbf{increase by one}$ the number of connected components of the (sub)level set of a Morse function. More precisely, passing an index zero critical point as the effect of taking the disjoint union on a $n$-ball (resp. $(n-1)-$ball) with the sublevel set (resp. level set) of the Morse function,

2) each time you pass an index $\lambda=1$ critical point you $\textbf{can increase by one}$ the number of connected components of the (sub)level set of your Morse function. This is because the attaching region of a one handle is disconnected. Specifically, the attaching region of a one handle consists of two $n$-balls called the $\textit{feet}$ of the one handle. If the two feet of the one handle you are attaching while you pass an index one critical point are on two different connected components $A$ and $B$, after you pass the critical point (and hence you glue the one handle) you will get the boundary connected sum $A \natural B$ on the sublevel set. On the level set of your Morse function right after the critical point this operation will reflect in the connected sum $\partial A\# \partial B$. In the case when the feet of the one handle are on the same connected component of the sublevel set right before the critical point, going through the critical point will not effect the number of connected components of the sublevel set,

In your setup you are assuming that $W$ is connected and that the Morse function $f$ has only critical points of intermediate index, thus nothing can effect the connectedness of the (sub)level sets of $f$.

Hope this answer you question!

PS: in order to analyse what happens for index $\lambda= n-1, n$ critical points just change $f$ with $-f$. By doing this you exchange index $i$ critical points with index $n-i$ critical points, and you turn upside down the handle decomposition.

  • $\begingroup$ The number of components of a level set can certainly increase when you pass a critical point of index $n-1$ and always decrease when you pass a critical point of index $n$, and there is certainly no way for the number of components to increase when passing a critical point of index $1$. $\endgroup$ Nov 22, 2016 at 1:28
  • $\begingroup$ Yes you are right, I edited. $\endgroup$ Nov 22, 2016 at 1:39
  • $\begingroup$ This still isn't quite right. The point isn't just that the attaching region is connected, but also that (1) (each component of) the level set minus the attaching region is connected and nonempty (2) the glued on handle itself is also connected. Hence the effect of the surgery does not add any components (We are just gluing on some connected thing to some other already connected thing). This is because the attaching region is a trivial tubular neighborhood of a (k-1)-sphere (with $k >1 $ and importantly codim $>1$!) and in that case the complement is connected. $\endgroup$ Nov 22, 2016 at 3:30
  • $\begingroup$ A little bit pedestrian, but needed. Btw I don't think I stated something wrong. $\endgroup$ Nov 22, 2016 at 5:18

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