How to prove that two curves are not path homotopic I have a unit circle around origin.And another unit circle around $(2,0)$. Consider the domain $R^2 / \{(0,0)\}$. I am able to clearly see that both are not homotopic but i am unable to prove it rigorously.
We have to show that there exists no continuous deformation . So if we assume that deformed path at time t to be $\gamma_t(s)$. Then we have to show that  $\gamma_t(s)$ passes through origin for some t. But I am unable to come up with some contradiction. If anyone can give a hint as  to how to proceed it would be great.Thanks
 A: You should be able to show that unit circle around $(2,0)$ is homotopic to a constant loop.
You should also be able to prove that homotopy is an equivalence relation,
and that all constant loops are homotopic to one another.
(What I mean is, you should be able to do these just using the definition
of homotopy.)
Thus, your question comes down to proving that the unit circle around the origin
is not homotopic to a constant loop in $\mathbb R^2 \setminus \{(0,0)\}.$
Again, just using the definition, you could prove that this is equivalent to
showing that the unit circle is not homotopic to a constant loop regarded as curves in the unit circle.  (In other words, replacing $\mathbb R^2 \setminus \{(0,0)\}$ by the unit circle; one says that these two spaces are homotopy equivalent.)
This last statement can be a bit tricky to prove with absolutely no theory to help you, but if you keep reading, you will soon learn how to do it.  The easiest approach is to use the map $\mathbb R \to \text{unit circle}$ given by $t \mapsto e^{2\pi it}$ (a special example of a universal covering map).  It lets you make precise the intuitions you probably have about whether or not a loop winds non-trivially around the unit circle. 
