Entropy of a North South Transformation. Let $f:\mathbb{S}^2\to\mathbb{S}^2$  be a continuous north south Transformation, in other words, the point $(0,0,1)$ is a global attractor for $f$  and $(0,0,-1)$  is a global attractor for $f^{-1}$.  
How  to calculate  the entropy of $f$?
I do not want to use the theorem:
$$h_{top}(f)=h_{top}(f|NW).$$
Thanks in advance!
 A: One easy way to do this is to use the variational principle to say that the topological entropy of $f$ is the supremum over all $f$-invariant probability measures $\mu$ of the metric entropy $h_\mu(f)$, that is $$h_{top}(f) = \sup_\mu h_\mu(f).$$ The problem then becomes a matter of understanding $f$-invariant probability measures. Suppose $\mu$ is an $f$-invariant probability measure. Then the Poincaré recurrence theorem implies that almost every point in the support of $\mu$ is recurrent for $f$. Since the only points in $\mathbb{S}^2$ which are recurrent for $f$ are the north and south pole, any such $\mu$ must be supported on these two points, i.e., $\mu = a\delta_{N} + b\delta_S$, where $a,b\geq 0$, $a + b = 1$, and $\delta_N$ and $\delta_S$ denote the Dirac probability measures at the north and south poles, respectively. It is easy to check that $h_\mu(f) = 0$ for any such $\mu$. Thus $h_{top}(f) = 0$.
A: A simple way for this proof is see that any 2 power of the nort-south maps are conjugated, so fo every $n\in\mathbb{N}$ we have, 
$$h_{top}(T)=h_{top}(T^n)=n\cdot h_{top}(T)  $$
then we have $h_{top}(T)=0.$
