# Fixed points and their continuity (2)

In summary, the question was,

Let $f : I^2 \to I$ be a continuous map, where $I := [0,1]$ is the unit interval. It is a basic fact that for each $y\in I$, the function $x \mapsto f(x,y)$ admits a fixed point. I want to ask whether one can always choose those fixed points as a continuous function of $y$.
Question: Does there always exist a continuous path $\gamma : I \to I$ such that $f(\gamma(y),y) = \gamma(y)$ for every $y\in I$?

Noam D. Elkies gave a counter-example as follows.

Not necessarily. Let $f(x,0)=0$, $\,f(x,1/2)=x$, $\,f(x,1)=1$, and define $f(x,y)$ for $0<y<1/2$ and $1/2<y<1$ by linear interpolation (that is, $f(x,y) = 2yx$ and $f(x,y)=1-2(1-y)(1-x)$ respectively). Then $g(y)$ would have to be $0$ for $y<1/2$ and $1$ for $y>1/2$, so $g$ cannot be continuous at $y=1/2$.

In that example,$f(x,1/2)$ has more than one fixed point with respect to $x$. Here a new question rises: if the function $x \mapsto f(x,y)$ admits a SINGLE fixed point for each $y\in I$, does the continuous path $\gamma$ exist?

Thank you.

I think, the answer is "yes". In your settings, the graph of $\gamma$ is the closed set $\{(y,x)\,|\,f(x,y)=x\}$, and, at least for compact Hausdorff spaces, a map is continuous iff its graph is closed (an easy exercise).
• Wait... Could you please elaborate on "set $\{(y,x)|f(x,y)=x\}$ is closed"? – LSZ Jan 28 '14 at 16:59
• This is $g^{-1}(0)$, where $g(x,y):=f(x,y)-x$. Informally, everything given by "continuous" equations and nonstrict inequality is closed: this is almost the definition of continuity. – Alex Degtyarev Jan 28 '14 at 18:02
• Try formally here. Suppose that $A=\{(y,x)|f(x,y)=x\}$ is not closed, then there exist a $(x',y')$ which is not in $A$ but close to $A$, such that $g(x',y')=f(x',y')-x'\neq 0$. Therefore, $g(x,y)$ is not continuous at $(x',y')$, which contradicts with the fact that $g(x,y)$ is continuous within $I^2$. – LSZ Feb 6 '14 at 11:15