I refer to the top answer on the following post: No diffeomorphism that takes unit circle to unit square.
If we assume that there is a diffeomorphism $f: \mathbb{R}^{2} \to \mathbb{R}^{2}$, we want to show that the image of the open unit ball can't be the open unit square. The strategy mentioned in the answer I've linked above is to show that this map (if it existed) should smoothly extend to the boundary, i.e. map the boundary of the open unit ball to the boundary of the square; however, we run into a problem at the corners of the square since the tangent vectors are perpendicular there, which implies that the differential (which we assume exists) must be zero, which contradicts our diffeomorphism assumption.
Can someone explain to me what tangent vectors in this context are, why they're perpendicular here, and why this implies that the differential is zero.
Look at the point $(1,1)$ on the upper right-hand edge of the square. If $f:\mathbb R^2\to \mathbb R^2$ is such a diffeomorphism, consider the map $\gamma:t\mapsto e^{2i\pi t}$ where we assume without loss of generality that $f\circ \gamma(0)=(1,1)$ and that there is an $\epsilon>0$ such that $f\circ\gamma(-\epsilon,0)\subseteq \{(1,y):0<y<1\}$ and $f\circ\gamma(0,\epsilon)\subseteq \{(x,1):0<x<1\}.$
Then, $t\in (-\epsilon,0)\Rightarrow f\circ\gamma(t)=(1,\gamma_1(t))$ for some smooth $\gamma_1$ and $t\in (0,\epsilon)\Rightarrow f\circ\gamma(t)=(\gamma_2(t),1)$ for some smooth $\gamma_2$. Then we have, $(f\circ \gamma)'(t)=(0,\gamma_1'(0))$ and $(f\circ \gamma)'(t)=(\gamma_2'(0),0)$ which means on taking right and left hand limits, that $(f\circ \gamma)'(0)=0$. But now we have a contradiction because $df$ is an isomorphism and $\gamma'(0)\neq 0.$