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Yesterday I asked a question about fixed point. Here is the link.
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.

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2 Answers 2

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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).

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  • $\begingroup$ Wait... Could you please elaborate on "set $\{(y,x)|f(x,y)=x\}$ is closed"? $\endgroup$
    – LSZ
    Jan 28, 2014 at 16:59
  • $\begingroup$ 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. $\endgroup$
    – Alex Degtyarev
    Jan 28, 2014 at 18:02
  • $\begingroup$ 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 $I^2$. $\endgroup$
    – LSZ
    Feb 6, 2014 at 11:15
  • $\begingroup$ @Lansiz: Well, doing serious mathematics, you should forget calculus 101. A function (map) is continuous iff the pull-back of any closed set is closed. This proves pretty much everything. BTW, the "easy exercise" that I mention in my post is easy assuming this definition; another crucial observation needed is that a continuous map from a compact space to a Hausdorff one is closed, i.e., it takes closed sets to closed sets. Without the compactness assumption the characterization of continuity in terms of graphs is obviously wrong. $\endgroup$
    – Alex Degtyarev
    Feb 6, 2014 at 11:56
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The question when and to what extent there is a continuous/computable map from the parameter to the fixed point is the focus of our paper: http://arxiv.org/abs/1206.4809

In our terminology, finding fixed points is a weakly computable operation, and a weakly computable operation with unique solutions is already computable (hence in particular continuous). Any other restriction on the fixed points will probably not suffice, as "closed choice for {0,1}" will be reducible to it.

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