Surjectivity of derivative of a vector valued function Let $f:\mathbb R^3\to \mathbb R^3$ be a function such that $f(x,y,z)=f(x+y,0,x+z)$ for all $(x,y,z)\in \mathbb R^3$. I want to prove that $f^{'}(x)$ can never be onto for all point $x\in \mathbb R^3$ of the type $(a,0,c)$.
Here if $f$ is a $C^{1}$ function, then by using inverse mapping theorem the problem may be solved. But unfortunately it is not given. Please help me to solve the problem.
 A: We have $f = f\circ T,$ where $T$ is the linear map $(x,y,z) \to (x+y,0,x+z).$ If we asssume $Df(p)$ exists for all $p,$ then by the chain rule
$$Df(p) = Df(T(p))\circ DT(p) = Df(T(p))\circ T,$$
where we have used the fact that $DT(p) = T$ (true for any linear map). Now the range of $T$ is contained in the two dimensional subspace $ \mathbb R \times \{0\} \times \mathbb R,$ so the rank of $T$ is $\le 2$ (it's actually $2,$ but never mind). Thus the rank of $Df(T(p))\circ T$ is at most $2,$ hence it is not surjective.
Like @Alex M., I do not understand the $(a,0,c)$ business.
A: Since you reference $f'$, $f$ must be differentiable. The inverse function theorem can actually be relaxed from $C^1$ functions to differentiable functions (see here), so the argument you had in mind still applies.
A: Remember that $f'(x,y,z) : \Bbb R^3 \to \Bbb R^3$ is a linear map; to say that it is surjective means, in this case, to say that $\text{rank } f'(x,y,z) = 3$, which is equivalent here to saying that $\det f'(x,y,z) \ne 0$. Letting $f = (u,v,w)$, let us compute this determinant.
Begin by computing the partial derivatives of the given relationship:
$$\begin{align}
\frac {\partial f} {\partial x} (x,y,z) = & \frac {\partial f} {\partial x} (x+y, 0 ,x+z) + \frac {\partial f} {\partial z} (x+y, 0, x+z) \\
\frac {\partial f} {\partial y} (x, y, z) = & \frac {\partial f} {\partial x} (x+y, 0, x+z) \\
\frac {\partial f} {\partial z} (x, y, z) = & \frac {\partial f} {\partial z} (x+y, 0, x+z)
\end{align}$$
and replace the second and third pieces of relationship into the first one, to obtain
$$\frac {\partial f} {\partial x} (x,y,z) = \frac {\partial f} {\partial y} (x, y, z) + \frac {\partial f} {\partial z} (x, y, z) ,$$
and this is valid at evry point of $\Bbb R^3$, not just at points of the form $(a,0,c)$!
Writing the components of $f$ explicitly and dropping the argument $(x,y,z)$, the above means that
$$\begin{align}
\frac {\partial u} {\partial x} = & \frac {\partial u} {\partial y} + \frac {\partial u} {\partial z} \\
\frac {\partial v} {\partial x} = & \frac {\partial v} {\partial y} + \frac {\partial v} {\partial z} \\
\frac {\partial w} {\partial x} = & \frac {\partial w} {\partial y} + \frac {\partial w} {\partial z}
\end{align}$$
(again, this is valid everywhere).
From this it follows that
$$\det f'(x,y,z) = \begin{vmatrix}
\frac {\partial u} {\partial x} & \frac {\partial u} {\partial y} & \frac {\partial u} {\partial z} \\
\frac {\partial v} {\partial x} & \frac {\partial v} {\partial y} & \frac {\partial v} {\partial z} \\
\frac {\partial w} {\partial x} & \frac {\partial w} {\partial y} & \frac {\partial w} {\partial z}
\end{vmatrix} (x,y,z)
= \begin{vmatrix}
\frac {\partial u} {\partial y} + \frac {\partial u} {\partial z} & \frac {\partial u} {\partial y} & \frac {\partial u} {\partial z} \\
\frac {\partial v} {\partial y} + \frac {\partial v} {\partial z} & \frac {\partial v} {\partial y} & \frac {\partial v} {\partial z} \\
\frac {\partial w} {\partial y} + \frac {\partial w} {\partial z} & \frac {\partial w} {\partial y} & \frac {\partial w} {\partial z}
\end{vmatrix} (x,y,z) = 0$$
because the first column is a linear combination of the other two. Since the determinant is $0$, it follows that $\text{rank } f'(x,y,z) \le 2$, so $f'(x,y,z)$ is not surjective for any $(x,y,z) \in  \Bbb R^3$ (not just for the points of the form $(a,0,c)$!).
(In my opinion, this is a very poor exercise because it suggests the student that the points $(a,0,c)$ are in some way special and that the statement of the problem holds only for them, which is clearly not the case. I bet there are better course materials available.)
A: I really like zhw.'s answer but let me show another approach.
For a fixed $(x,y,z)$ define a function
$$
h(t) = f(x+t, y-t, z-t), \quad  t \in \mathbb R.
$$
Obviously, $h(t) = f(x+y,0,x+z)$, so it's constant and if a derivative of $f$ exists at $(x,y,z)$, we have
$$
0 = h'(0) = f'(x,y,z) (1,-1,-1).
$$
Thus $f'(x,y,z)$ is not surjective.
We don't even need the existence of a Frechet derivative. If a linear Gateaux derivative at $(x,y,z)$ exists than it can't be surjective either.
