I have the wave equation, $u:\mathbb{R}^2\rightarrow\mathbb{R}, c\in\mathbb{R}_{\neq 0}$, \begin{gather} \frac{\partial^2 u(x,t)}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 u(x,t)}{\partial t^2} \end{gather} I know the general solution is $u(x,t)=f(x+ct)+g(x-ct)$.
We can show this with the change of variables, $\xi =x+ct$ and $\eta =x-ct$, so $u(\xi, \eta)=f(\xi)+g(\eta)$. And from here it is the chain rule. So far so good.
However, I don't grasp the change of variables. Why can we write $ u(\xi, \eta)=f(\xi)+g(\eta)$? I mean, If I have $$u(x,t)=f(x+ct)+g(x-ct) \tag 1$$ and introduce $\xi =x+ct$ and $\eta =x-ct$, I solve for $x$, i.e. \begin{align} \xi &=x+ct \quad \iff \quad t=\frac{\xi - x}{c} \\ \eta &=x-ct \quad \iff \quad t=\frac{x-\eta }{c} \end{align} \begin{equation} \boxed{x=\frac{\xi +\eta}{2}} \tag 2 \end{equation} And same for $t$, so \begin{equation} \boxed{t=\frac{\xi -\eta }{2c}} \tag 3 \end{equation}
I now change variables in $(1)$, i.e. \begin{align} u\bigg(\frac{\xi+\eta}{2}, \frac{\xi-\eta}{2c}\bigg)=f(\xi)+g(\eta). \tag 4 \end{align} Why is this wrong? Why is $u(x,t)=u(\xi,\eta) \tag 5$ and not $u(x,t)=u\bigg(\frac{\xi+\eta}{2}, \frac{\xi-\eta}{2c}\bigg)?\; \tag 6$
Because now we are considering the $u$ as a funcion of the two new variables then
$$u(x,t)=u\bigg(\frac{\xi+\eta}{2}, \frac{\xi-\eta}{2c}\bigg)=f(\xi)+g(\eta)=u(\xi,\eta)$$
The confusion arises because the same symbol is used for $u$ but it is justified since, even if the expressions for the two function are different, they represent the same function.