Elementary Morse theory

Consider the restriction $f$ of the map $F:\mathbb R^3\rightarrow \mathbb R,(x,y,z)\mapsto z$ to $S^2$ : indeed it is a smooth function and according the method of Lagrange multiplier, you know that $f$ has only two critical points (no choice, it has to be the max and min of $f$) located at $p_\epsilon=(0,0,\epsilon)$, $\epsilon=\pm 1$.

To find the nature of this two points, you need to find a chart around $p_\epsilon$ and compute the Hessian of $f$.

Around $p_\epsilon$, the map $\varphi : (x,y)\in B(0,r) \mapsto (x,y,\epsilon\sqrt{1-x^2-y^2})$ is nice chart and $f\circ \varphi (x,y)=\epsilon\sqrt{1-x^2-y^2}$ on a neighbourhood of $p$. The study of the map $g=f\circ \varphi$ around $(0,0)$ will give you the nature of $p_\epsilon$.

You can check that $$\dfrac{\partial^2 g}{\partial x^2}(0,0)=\dfrac{\partial^2 g}{\partial y^2}(0,0)=-\epsilon$$ and $$\dfrac{\partial^2 g}{\partial x \partial y}(0,0)=0.$$

It means that the Hessian of $f$ at $p_\epsilon$ is either negative definite or positive definite according to $\epsilon =1$ or $\epsilon =-1$. In other words $f$ looks like $(x,y)\mapsto f(p_\epsilon)-\epsilon(x^2+y^2)$ around $p_\epsilon$ where $(x,y)$ are local coordinates and the number of negative square is the Morse index (according to Morse lemma).

So for $\epsilon =1$, $p_\epsilon=(0,0,1)$ is a maximum and its index is equal to $2$. For $\epsilon = -1$, $p_\epsilon=(0,0,-1)$ is a minimum and its index is equal to $0$.

Finally, if I denote by $C_i$ the $\mathbb Z/2\mathbb Z$-vector space generated by critical points of $f$ of index $i$, the Morse theory tells you that we have an exact sequence $$\{0\}\rightarrow C_2 \rightarrow C_1 \rightarrow C_0 \rightarrow \{0\}.$$

In this case, $C_0\simeq \mathbb Z/2\mathbb Z \simeq C_2$ and $C_1=\{0\}$, so the Morse cohomology of this complex is $$HM_i=\left\{\begin{array}{ll} \mathbb{Z} / 2 \mathbb Z & \text{if } i=0,i=2 \\ \{0\} & \text{otherwise}\end{array}\right.$$ and $\chi(S^2)=1-0+1=2$.

To show that a compact manifold $V$ of dim $d$ has a lot of Morse functions, one can use the map $f_p(x)=\|x-p\|^2$ (the square gives regularity). Let's think of $V$ as a submanifold of $\mathbb R^n$.

• Check that a point $c\in V$ is a critical point of $f_p$ iff $c-p\perp T_cV$.
• If $v=v(x_1,\cdots,x_d)\in V$ is a local chart around $c$, then $$\dfrac{\partial^2 f}{\partial x_i\partial x_j}=2(<\dfrac{\partial v}{\partial x_i},\dfrac{\partial v}{\partial x_j}>+<v-p,\dfrac{\partial^2 v}{\partial x_i\partial x_j}>).$$ You want to show that for almost every $p$, this matrix is definite i.e. of rank $d$.
• Consider the normal bundle $N=\{(v,w)\in V\times \mathbb R^n | w\perp T_xV\}\subset V\times \mathbb R^n$ and $F:N\rightarrow \mathbb R^n, (v,w)\mapsto v+w$. $N$ is a manifold of dimension $n$ and $F$ is smooth.
• Check that $p=v+w\in \mathbb R^n$ is a regular value of $F$ iff $\forall v,w$ s.t. $p=v+w$ the matrix $M=(m_{ij})$ given by $$m_{ij}=<\dfrac{\partial v}{\partial x_i},\dfrac{\partial v}{\partial x_j}>+<w,\dfrac{\partial^2 v}{\partial x_i\partial x_j}>$$ is invertible.

With this four points checked, apply the Sard lemma and you get the proof.

Transversality helps you to construct in generic way the morse complex. Roughly speaking, to define the complex, you need a nice morse function, a nice pseudo-gradient vector field, etc... that all fit together. It is not obvious that one can always do that but the transversality conditions allows you to say : if it doesn't work at some point, then you can perturb a little bit all this tools so that it will.

A couple of years ago, Alexander Ritter taught a course in Morse homology whose lectures notes can be found here. They take a modern perspective, and so are quite heavy on functional analysis.

In any case, there should be something of use there, in particular, they answer your questions about independence of choice of Morse function and where transversality comes into the picture.