Where's the error in my "proof" about iterated multivariable functions? I've tried to prove a property of a kind of iterated multivariable functions I am playing with and I thought I did it, but now I've found a lot of counter examples.
I'm still going to school, but I hope the formalism isn't to horrible.
I started by defining injectivity, surjectivity and bijectivity with respect to a single variable for multivariable functions:
Def: A funtion $f(x,y):\mathbb{R}²\to\mathbb{R}$ is injective/surjective/bijective  with respect to $x$ if and only if $f_y(x)=f(x,y)$ is injective/surjective/bijective for all $y\in\mathbb{R}$.
I thought that I proved the Proposition that $f(x_0,y_0)=x_0$ was true if $f(f(x_0,y_0),y_0)=x_0$ is true and $f$ is bijective w.r.t. x and continuous, but I found some counter examples.
One of them is $f(x,y)=\cos(y)-x³$. I wanted to illustrate this with some pictures, but sadly I haven't gotten any reputation yet, so I can't post any.
First I "proved" a Lemma:
If the the system of equations $$f(x)=y\\f(y)=x$$ has any solutions and $f$ is continuous, x=y is true for at least one of them (this seems kind of obvious, but I couldn't find a proof online).
Proof by contradiction:
assume $f(x)\neq x$ for all $x\in\mathbb{R}$.  It follows from continuity that $f(x)<x$ for all $x\in\mathbb{R}$ or $f(x)>x$ for all $x\in\mathbb{R}$.  For $f(x)<x$ this means that $f(f(x))<f(x)<x$, but the original system of equations means that $f(f(x))=x$.  This is a contradiction.  The proof is exactly the same for the case $f(x)>x$.  this proves the Lemma.
Now I tried to prove the Proposition.
Proof:
Assume $f$ is bijective w.r.t $x$ and continuous and $f(f(x_0,y_0),y_0)=x_0$.  It follows from bijectivity that there is exactly one $x':=f(x_0,y_0)$ that solves this with $f(x',y_0)=x_0$. Fix $y_0$ and set $f_y(x)=f(x,y_0)$. 
Then we have $f_y(x_0)=x'$ and $f_y(x')=x_0$.
 It follows from the Lemma that $x_0=x'$ for at least one solution, but since $f_y$ is bijective this is the only solution.
 This means that $f(x_0, y_0)=x_0.$  q.e.d.
But as I said earlier there are some counter examples, though it does seem like $f(x_0, y_0)=x_0$ is really true when $f$ is bijective w.r.t. $x$ and continuous and $f(f(f(x_0,y_0),y_0),y_0)=x_0$, though I haven't proved that yet.
Thank you for your help in advance.
 A: What you proved in your lemma is that if there exist $x, y \in \mathbb{R}$ such that $f(x) = y$ and $f(y) = x$ for a continuous function $f: \mathbb{R} \to \mathbb{R}$, then there exists $z \in \mathbb{R}$ such that $f(z) = z.$
It is in general not true that $z = x = y$, because e.g. for $f(x) := -x$ we have $f(1) = -1$ and $f(-1) = 1$, but we only have $f(0) = 0$.
In your proof of the proposition you have that $f_y(x_0) = x'$ and $f_y(x') = x_0$, so you can apply your lemma and deduce that there exists $z \in \mathbb{R}$ such that $f_y(z)=z$. But you can not infer from your lemma that this $z = x_0 = x'.$ This is the (or at least one) error in your proof.
What is in my opinion worthwhile to think about is why this error happened. I reckon this is because you phrased the existence of a $z \in \mathbb{R}$ with $f(z) = z$ in your lemma as "$x=y$ is true for at least one of them", which is not a particularly precise mathematical statement. I recommend taking a look at how to quantify variables using logical quantifiers.
A: Consider the plane $ f(x, y) = x + y $
If we fix $y=0$, we get the familiar "line through the origin" and indeed, there are solutions to $ f(f(x, y), y) = x$, infact, any x satisfies this (remember, $y = 0$ fixed) since $ f(f(f...(x))...), 0) = x $
The problem is the last line, where you sate "but since $ f_y $ is bijective, this is the only solution" the function I had above was also bijective, yet there were infinitely many solutions? what happened is that you fixed two things to each other that weren't fixed by anything else, and so weren't actually fixed at all (don't want to spoil it all, maybe you want to explore it bit more, if you want a more descriptive answer though, let me know)
