I am studying Morse theory and I am bit confused about what happens to a manifold when going through a critical value of a Morse function.
Here is the setting: $M$ is a compact manifold and $f : M \to \mathbb{R}$ is a Morse function. Assume $c$ is a critical value such that there is one only one critical point in $f^{-1}[c-\epsilon, c+ \epsilon]$. Then, $M_{c+ \epsilon}$ is obtained by attaching a handle to $M_{c - \epsilon}$.
In the above, $M_a = f^{-1}[- \infty, a]$. What I do not get is the case where the index of the critical point $p$ is equal to the dimension of $M$. In that case, there's a local maximum at $p$, but it might not be global. What I thought one can do in this case is replace $M$ by $M_{c+ \epsilon}$, but there might be the need to "stretch" $M_{c- \epsilon}$ somewhere to get $M_{c+ \epsilon}$.
Is what I am saying correct? The reference I am reading is "Differential Topology" by Amiya Mukherjee.
I am not sure what the issue is here. When $p$ is an index $n$ critical point, you are attaching an $n$-cell. From a Morse theoretic perspective, it doesn't matter whether the top index critical point is a global maximum or not, since either way, you are attaching an $n$-cell. Note that these cell attachments are up to homotopy, maybe you are confused about this?
Let me try to give you an example. Consider our manifold as an island with two mountains, one is higher than the other. When the Morse height function goes through the peak of the smaller mountain, the homotopy type of $M_a$ changes by attaching a $2$-cell to the smaller mountain. If you consider what happens to the other mountain (and as far as I understand, this is the part you are confused about), the homotopy type of the bigger mountain does not change. You can simply deformation retract it to its older form. Let's say the peaks of the smaller and bigger mountains are $p$ and $q$, respectively. Then, below $p$, you are missing the peaks of both mountains; between $p$ and $q$, you are only missing the peak of the bigger mountain; and above $q$, you have the entire island. Does this answer your question? In short, there is no 'stretching' involved. For large enough $a$, the set $M_a$ will be $M$ itself, so you don't need to try involve $M$ somehow.