I encountered this issue recently, but do not know of any general results to deal with it, so I would appreciate any pointers. 



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> Let $\mathbb T=\{z\in\mathbb C\mid |z|=1\}$, and let $f:\mathbb T\to\mathbb C$ be continuous and injective, so its image $\mathbb T'$ is a Jordan loop. Under what (general) conditions can we ensure that there is a homeomorphism between the unit disc and the interior of $\mathbb T'$ whose extension to the boundary is $f$? 
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Moreover, if there are reasonable conditions that ensure this, and $f$ is $C^\infty$, can we further require some nice regularity (perhaps even $C^\infty$) of the extension as well?



Extending Jordan loops
What you're asking is equivalent to asking whether any homeomorphism $g : S^1 \rightarrow S^1$ can be extended to a homeomorphism of the disc. This is easy -- write the disc in polar coordinates $(t,\theta)$ with $\theta \in S^1$, and define an extension $G(t,\theta) = (t,g(\theta))$.


The question about whether this can be done smoothly if $g$ is smooth is more subtle. Observe that the above also works for $S^k$ with $k > 1$. The smooth version fails in higher dimensions and is responsible for the existence of exotic spheres. However, for $k=1,2$ there is no problem. For $k=1$, this is a theorem of Smale; see


Smale, Stephen 
Diffeomorphisms of the 2-sphere. 
Proc. Amer. Math. Soc. 10 1959 621–626.


For $k=2$, it is a much deeper theorem of Hatcher; see


Hatcher, Allen E.
A proof of the Smale conjecture, Diff(S3)≃O(4). 
Ann. of Math. (2) 117 (1983), no. 3, 553–607. 


