Manifold structure of 2-sphere: Use of inversions as transition functions I have a $S^2$ sphere. We proved it's a manifold it by showing that it could be covered by 2 coordinate charts through stereographic projections.
$\psi$ is the projection from the north pole
$\phi$ is the projection from the south pole
We found a line $(r(t)=<at,bt,1-t))$ that connects the north pole to the $(a,b,0)$ plane. and computed it's intersection $(\psi^{-1}(a,b)= (\frac {2a}{a^2+b^2+1},\frac {2b}{a^2+b^2+1}, \frac {a^2+b^2-1}{a^2+b^2+1}$)
we did something similar to find $\phi$ and then composed $(\phi \circ \psi^{-1}(a,b)=(a/(a^2+b^2), b/(a^2+b^2))$
These trasition maps are just inversion about the origin in the (a,b)-plane. (with the origin removed) and it is a $c^\infty$ function. 
Can someone explain to me what this bold statement exactly means?Also,  How do we know the origin is removed?
 A: An inversion is just a type of geometric transformation! In plane polar coordinates, an inversion sends $(r,\phi)$ to $ (1/r, \phi)$. If you write this transformation in terms of the cartesian coordinates $(a,b)$, you'll see that this is the same thing as the transition function $\phi\circ \psi^{-1}$ that you wrote down in the question.
But for the purposes of understanding that your chart on $S^2$, this business with the "inversion" is not important at all! What is important is that your transition function $\phi\circ \psi^{-1}$ is smooth. The transition function is your prescription for how you switch from the coordinates you use to parametrise the patch containing the north pole to the coordinates you use to parametrise the patch containing the south pole. This rule for changing coordinates had better be given by a smooth function, otherwise your $S^2$ won't be a smooth manifold.
The reason why the origin $(a,b) = (0,0)$ is excluded from the domain of your transition function is that the origin $(a,b) = (0,0)$ is precisely the north pole of $S^2$. Remember, the north pole is only covered by your first coordinate chart. The south pole is only covered by your second coordinate chart. Every other point is covered by both charts. So your prescription for "switching coordinates" is not needed at the north pole; it is only needed in between the poles. (Besides, your transition function is not well-defined at the north pole.)
