Yesterday I asked a question about fixed point. Here is the [link](https://mathoverflow.net/questions/155553/fixed-points-and-their-continuity).  

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](https://mathoverflow.net/users/14830/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.



Fixed points and their continuity (2)
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).


