Let $f:[0,1]^n \rightarrow [0,1]^n$ be a continuous mapping. Brouwer's fixed point theorem says that $f$ has a fixed point, i.e., some $x$ such that $f(x) = x$.

Suppose we have a continuous family, i.e., a continuous function $f:[0,1]^n \times [0,1] \rightarrow [0,1]^n$. Then for each $r \in [0,1]$ we have that there is a point $x_r$ such that $f(x_r,r) = x_r$. It is known that, in general, there is no continuous mapping $r \mapsto x_r$. The following counter-example exists in the case $n=1$: take $f(x,r) = 2rx$ if $r \leq 1/2$ and $f(x,r) = (2r-1) + (2-2r)x$ - for $r < 1/2$ we have the only fixed point as 0, and $r > 1/2$ we have the only fixed point as $1$.

My question is "what happens if we demand there are unique fixed points?" i.e., if every $f(\cdot,r)$ has a unique fixed point, is the mapping $r \mapsto x_r$ continuous?

In the case $n=1$ my intuition (which is very possibly wrong!) seems to indicate that if each of the functions $f(\cdot,r)$ has a unique fixed point, then the mapping sending such a function to its fixed point is continuous.

I'm sure this has been studied before by someone, however searching for it is a little difficult. Does anyone have any references, or know what happens for other $n$? I have little to no geometric insight as to what happens for $n \geq 2$.