(My question is related to the Brouwer fixed-point theorem.)
Let $B$ be a closed ball of $\mathbb{R}^n$.
Q 1. If $f : B \rightarrow B$ is a continuous function, then is there a homotopy between $id_B$ and $f$?
Assume the answer of the above question is yes. Then we can denote $\{f_r\}$ be a family of continuous functions where $0 \leq r \leq 1$ such that $\{f_r\}$ makes the homotopy between $id_B$ and $f$. By the Brouwer fixed-point theorem, the set $F_r := \{x ~|~ f_r(x) = x \}$ is nonempty for every $0 \leq r \leq 1$.
Q 2. Is there a continuous function $g: [0,1] \rightarrow B$ satisfying $g(r) \in F_r$ for every $0 \leq r \leq 1$? In other words, is there a path $g$ which passes through fixed points in the homotopy?
Thank you for your interest.
(1) is obviously true: the straight line homotopy $f_t=tf+(1-f)\operatorname{id}$.
(2) doesn't work. Here is a sketch: Let $x_0,x_1$ be distinct points, $f_0=f_{1/2}=f_1=\operatorname{id}$, $f_{1/4}=x_0$, $f_{3/4}=x_1$ and fit in the homotopies in (1). All nonidentity $f_r$ here only have $x_0$ or $x_1$ as the fixed point. So your $g$ must jump at $r=1/2$.