checking whether functions satisfy Inverse Function Theorem. I've my exam tomorrow and this question is expected  to come but donot know how to solve...
Here's the INVERSE FUNCTION THEOREM stated in my notes:
It says:

Let $E\subseteq \mathbb R^n$ be open  and $f:E\to \mathbb R^n$ be a $C^1$ map.
  Suppose that for some $a\in E$,the linear map $f'(a)$ is invertible, and $b=f(a)$.Then
  
  
*
  
*there are open set $U$ and $V$ in $\mathbb R^n$ ,s.t. $a\in U,b\in V$ and $f|_U$ is $1-1$ and onto V,that is $f(U)=V$  
  
*If g is inverse of $f|_U$ ,so $g:V\to U$ and $g(f(x))=x$ for all $x\in U$,then $g\in C^1(V,U)$  

I had to check whether following satisfy the above hypothesis of inverse function theorem on $D$:  
$1.)$ $g(x)=x+c$ , $D=\mathbb R^n$ ,
$2.)$ $g(s,t)=(s+2)e_2+(s-t)e_2$ , $D=\mathbb R^2$ ,
$3.)$ $g(s,t)=(s^2-t^2)e_2+(st)e_2$ , $D=\mathbb R^2$ \ ${(0,0)}$.
I think that $1.)$ satisfies all hypothesis as it has a inverse.... 
I don't know how to solve these questions...any hint...  
 A: All the functions given by the OP are $C^\infty$ smooth.


*

*The derivative of $g$ is the identity matrix and it is invertible.
It is a good exercise to prove that the derivative of a linear map is the map itself; that is, if $f(x)=Ax$ for a matrix $A$, then $f'(x)=A$ for all $x$.

*Now $g(s,t)=(g_1(s,t),g_2(s,t))=(0,2+2s-t)$.
This is again a linear mapping, but let us calculate the derivative in a more general way.
The derivative of this is
$$
g'(s,t)
=
\begin{pmatrix}
\partial_sg_1(s,t) & \partial_tg_1(s,t) \\
\partial_sg_2(s,t) & \partial_tg_2(s,t)
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 \\
2 & -1
\end{pmatrix}.
$$
Note that this matrix does not depend on $(s,t)$; this is because of linearity.
This is not an invertible matrix so the hypothesis of the inverse function theorem is not satisfied at any point.
(Also the conclusion is false since then $g$ would have to be an open mapping and it is not.)

*Now $g(s,t)=(0,s^2-t^2+st)$.
As above, we find
$$
g'(s,t)
=
\begin{pmatrix}
0 & 0 \\
2s+t & -2t+s
\end{pmatrix}.
$$
This matrix does depend on $(s,t)$ but it is never invertible.
If in the last example you meant $g(s,t)=(s^2-t^2,st)$ instead, you would get
$$
g'(s,t)
=
\begin{pmatrix}
2s & -2t \\
t & s
\end{pmatrix}.
$$
The determinant of this matrix is $2(s^2+t^2)$, which is nonzero in $D$.
Thus the hypothesis of the inverse function theorem is satisfied in all points $a\in D$.
