Q:

Rank of a cohomology group, Betti numbers.

How is the rank of a cohomology group computed and what does it convey?  I am trying to understand the concept behind betti numbers in a simplicial homology. 
Edited with details:
Given a set of nodes/vertices/points X, let $C_{l}(X)$ denote the subsets of $X$ with cardinality $|C_{l}(X)|=l+1$. $\partial_{l}$ and $\delta_{l}$ are bounded linear maps with $\partial_{l}$:$C_{l-1}(X)\rightarrow C_{l}(X)$  satisfying \begin{equation}\partial_{l-1}\circ\partial_{l}=0\end{equation} and are also known as the boundary$(\partial_{l})$and co-boundary operators($\delta_{l}$). Here, $\delta_{l}=\partial_{l}^{*}$ is the adjoint operator $\delta_{l}$:$C_{l}(X)\rightarrow C_{l+1}(X)$ with \begin{equation}\delta_{l}\circ\delta_{l-1}=0\end{equation}. The following is the definition of a co-boundary operator \begin{equation}(\delta_{l}\, f)(x_{0},x_{1},...,x_{l+1})=\sum_{i=0}^{l+1}(-1)^{i}f(x_{0},x_{1},...\hat{x_{i}}..,x_{l+1})\end{equation} and $\hat{x_{i}}$ indicates that it is omitted from the summation. Example to, Verify this: $$ f(s_{i},s_{j},s_{k},s_{l})= f(s_{j},s_{k},s_{l})-f(s_{i},s_{k},s_{l})+f(s_{i},s_{j},s_{l})-f(s_{i},s_{j},s_{k})$$ 
$$= f(s_{k},s_{l})-f(s_{j},s_{l})+f(s_{j},s_{k})-f(s_{k},s_{l})+f(s_{i},s_{l})-f(s_{i},s_{k})+f(s_{j},s_{l})-f(s_{i},s_{l})$$ $$+f(s_{i},s_{j})-f(s_{j},s_{k})+f(s_{i},s_{k})-f(s_{i},s_{j})=0 $$
I want a matrix representation of the coboundary and boundary operators. How is that representation done? I am not aware of the computational aspects of building a co-boundary operator in a matrix form from a simplicial complex. I believe if we have a matrix representation for one operator, the other would be its transpose.

A:

You are correct. Given a chain complex, and a specified basis for each chain group, consider the dual basis. Then the matrix of the coboundary operator is the transpose of the boundnary operator matrix. Also the rank of $H^i(X)$ is equal to the dimension of $H^i(X;\mathbb Q)$ as a rational vector space, and this is isomorphic to $H_i(X;\mathbb Q)$. So in fact the ranks of the cohomology groups are equal to the ranks of the corresponding homology groups, even though the torsion will shift around.

