A problem about differential of a smooth map Problem:
We have a map $ f: \mathbb{R}^3\rightarrow M(2,R)$
$$f(x,y,z)=exp \begin{pmatrix}
    x & y+z \\
    y-z & -x \\
    \end{pmatrix}$$
Here, we denote the set of all 2-dimensional real square matrices as $M(2,R)$.
Find the points where the differential of $f$ is not injective and draw the graph of this set.
What I have done:
(You can ignore the following part)
Let's denote $\begin{pmatrix} x & y+z \\ y-z & -x \\ \end{pmatrix}$ as $A$.
Then $A^2 = \begin{pmatrix} x^2+y^2-z^2 & 0 \\ 0 & x^2+y^2-z^2 \\ \end{pmatrix} = (x^2+y^2-z^2)I$
Then, we can get the value of $A^n:$
$$ 
\begin{array}{c|c|c}
  &\\\hline
  A^0&I&\\
  A^1&&A\\
  A^2&(x^2+y^2-z^2)I&\\
  A^3&&(x^2+y^2-z^2)A\\
  A^4&(x^2+y^2-z^2)^2I&\\
  A^5&&(x^2+y^2-z^2)^2A\\
\cdots&\cdots&\cdots
\end{array}
 $$
It is easy for us to calculate:
$$\begin{align}exp A 
& = I+ \color{red}{A}+\frac{A^2}{2!}+\color{red}{\frac{A^3}{3!}}+\frac{A^4}{4!}+\color{red}{\frac{A^5}{5!}}+\cdots \\  
& = I (1 + \frac{(x^2+y^2-z^2)}{2!} + \frac{(x^2+y^2-z^4)^2}{4!} +\cdots )+ \color{red}{A} ( \color{red}{1}+ \color{red}{\frac{(x^2+y^2-z^2)}{3!}} + \color{red}{\frac{(x^2+y^2-z^2)^2}{5!}} +\cdots ) \\ 
& = I (1 + \frac{(x^2+y^2-z^2)}{2!} + \frac{(x^2+y^2-z^4)^2}{4!} +\cdots )\\
& + \color{red}{\frac{A}{\sqrt{x^2+y^2-z^2}}} ( \color{red}{\sqrt{x^2+y^2-z^2}}+ \color{red}{\frac{(x^2+y^2-z^2)^\frac{3}{2}}{3!}} + \color{red}{\frac{(x^2+y^2-z^2)^\frac{5}{2}}{5!}} +\cdots ) 
\end{align}$$
When $x^2+y^2-z^2>0$,
$$\begin{align}exp A 
& = cosh\sqrt{x^2+y^2-z^2}I + \color{red}{\frac{A}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} \\
& = \begin{pmatrix}
    cosh\sqrt{x^2+y^2-z^2} + \color{red}{\frac{x}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} & \color{red}{\frac{y+z}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}}  \\
    \color{red}{\frac{y-z}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} & cosh\sqrt{x^2+y^2-z^2} - \color{red}{\frac{x}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}}  \\
    \end{pmatrix}
\end{align} \tag{1}$$
Similarly, We can calculate that when $x^2+y^2-z^2<0$,
$$exp A= \begin{pmatrix}
    cos\sqrt{z^2-(x^2+y^2)} + \color{red}{\frac{x}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}} & \color{red}{\frac{y+z}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}}  \\
    \color{red}{\frac{y-z}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}} & cos\sqrt{z^2-(x^2+y^2)} - \color{red}{\frac{x}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}}  \\
    \end{pmatrix} \tag{2}$$
When $x^2+y^2-z^2=0$, $ exp A = \begin{pmatrix} 1+x & y+z\\ y-z & 1-x\end{pmatrix} \tag{3}$
The standard way to solve the problem may be to identify $M(2,R)$ as $\mathbb{R}^4:$
$$\begin{pmatrix} x & z \\ y & w\end{pmatrix} \mapsto (x,y,z,w)$$
then calculate the differential for (1) (2) and (3) and decide when their rank$<3$.
But I found it is hard to calculate. So is to find under which condition rank $<3$.
Any one can give me some ideas or which books deal with related topics?

You can write $f = g \circ h$ where
$$\begin{array}{l|rcl}
g : & M(2, \mathbb R) & \longrightarrow & M(2, \mathbb R) \\
    & A & \longmapsto & \exp A \end{array}$$
and
$$\begin{array}{l|rcl}
h : & \mathbb R^3 & \longrightarrow & M(2, \mathbb R) \\
    & (x,y,z) & \longmapsto & \begin{pmatrix}
    x & y+z \\
    y-z & -x \\
    \end{pmatrix} \end{array}$$
According to the chain rule
$$f^\prime(x,y,z) = g^\prime(h(x,y,z)) \cdot h^\prime(x,y,z)$$ and $f^\prime$ is injective if and only if both $g^\prime(h(x,y,z))$ and $h^\prime(x,y,z)$ are injective. As $h$ is linear, its derivative is equal to itself at each point of $\mathbb R^3$. Also $g^\prime(A)= \exp A$.
The characteristic polynomial of $h(x,y,z)$ is $\mu_{h(x,y,z)}(X)= X^2 -(x^2+y^2-z^2)$. We need to deal with 3 cases, depending on the sign of $r(x,y,z) = x^2+y^2-z^2$.
Case 1: $x^2+y^2-z^2 = 0$
In that case, $\mu_{h(x,y,z)}(X)= X^2$, $0$ is an eigenvalue of $h$ (and $h^\prime$) and $f^\prime$ is not injective.
Case 2: $x^2+y^2-z^2 \gt 0$
In that case, $\mu_{h(x,y,z)}(X)$ has two distinct non vanishing roots and $h$ is diagonalizable. $g^\prime(h(x,y,z))$ is also diagonalizable and its eigenvalues are the exponential of the one of $h$. Hence, are also distinct and non zero. $g^\prime(h(x,y,z))$ is injective as $h^\prime(x,y,z)$ and finally $f^\prime(x,y,z)$ is injective too.
Case 3: $x^2+y^2-z^2 \lt 0$
$\mu_{h(x,y,z)}(X)$ has two distinct non vanishing pure imaginary roots $\sqrt{z^2 - x^2-y^2} i, -\sqrt{z^2 - x^2-y^2} i$ and $h$ is again diagonalizable. $g^\prime(h(x,y,z))$ is also diagonalizable and its eigenvalues are the exponential of the ones of $h$. Hence, are also distinct. $f^\prime$ is also injective in that case.
Finally, $f^\prime$ is not injective if and only if $(x,y,z)$ belongs to the cone $x^2+y^2 = z^2$.
