I'd like to show that any continuous mapping from a circle to a disc is homotopic to the trivial mapping. This is used in Coleman's proof that the homotopy group $ \pi ^1 ( S ^m ) $ is the trivial group (also known as why you can't ``lasso a baseketball'') he says that (see Aspect of Symmetry, pg 207),
Define a homotopy (from a circle to a disc) by shrinking the disc to its central point. You cannot lasso a basketball.
Equivalently Coleman says that any continuous mapping from a circle to a disc is the trivial mapping since you can always continuously deform the mapping to the identity. While this is somewhat intuitive since there are no topological defects, I'd like to show this explicitly using Coleman's "standard way" by deforming some mapping from $ t = 0 $ to $ t = 1 $: \begin{equation} ( r, \theta ) t + ( \text{identity mapping} ) ( 1 - t ) \end{equation} (where $r$ and $\theta$ denote functions of the angle on a circle to some point on the disc). Can somewhat clarify this point? In particular I'd like to know that the deformation that takes you from a generic mapping to the identity looks like.