Explain two different ways to geometrically interpret a map $ T: D^* \to D \ $ Explain two different ways to geometrically interpret a map $ \ T:D^{*} \to D \ $ where $ \ D^{*} , \ D \subset \mathbb{R}^2 \ $
, assuming
T
is
bijective and  $ \ C^{1} \ $. 
Answer:
We have to explain geometrically. 
Explanation  1:
Since $ \ T \ $ is bijective and $ \ C^{1} \ $ , $ \ T^{-1} \ $ is also bijective and $ \ continuous \ $.
Thus $ \ D^{*} \ $ is homeomorphic to $ \ D \ $
Thus $ \ D^{*} \ \ and \ \ D $ denotes the similar shape of figure  in a plane.
Am I right ?
If , then what would be the other way explanation.
 A: 
Since $T$ is bijective and $C^1$, $T^{-1}$ is continuous.

No. Consider $D^*=\{(x,y)\,:\, y=0\land x\ge 0$, $D=\{(x,y)\,:\, x^2+y^2=1\}$ and the map $\widetilde T:\Bbb R^2\to\Bbb R^2$ $$\widetilde T(x,y)=(e^y\cos(2\pi(1-e^{-x})),e^y\sin(2\pi(1-e^{-x})))$$ $\widetilde T$ is not bijective itself (otherwise it would indeed be a homeomorphism by invariance of domain). However, $T=\left. \widetilde T\right\rvert_{D^*}: D^*\to D$ is bijective, it's $C^1$ in any reasonable way, but its domain is not compact while its image is.

[$D$ and $D^*$ are homeomorphic and] thus $D$ and $D^*$ denote similar shape of figure in the plan.

Either tautologically - as in having defined "similar shape" as "being homeomorphic" - or rather arguably. For instance, if $C$ is Cantor's set, then  $C\times C$ is homeomorphic to $C$, while it isn't quite clear to me why the "shape" of $C\times\{0\}$ should be similar to the "shape" of $C\times C$.

What would be the other way explanation?

My opinion is that this isn't a real question, because it depends largely on information that is not available to anyone who isn't attending the same lectures you are.
