Is there a version for the Brouwer fixed point theorem for maps rather than spaces ? In other words, for a family of endomorphisms, can the fixed point be chosen continuously, under some assumptions ?

One way to make this question precise is as follows.

A standard formulation of the Brouwer fixed point theorem says that there is a fixed point of an endomorphism of a contractible compact space $K$, whenever the space is sufficiently "nice". In terms of maps, it means that the map $K\to \{pt\}$ is an acyclic fibration and a proper map.

Is there a version of the Brouwer fixed point theorem for "endomorphisms" of maps which are both acyclic fibration and a proper map ?

For example, the following naive version is false due to the counterexample below. Can one make stronger assumptions such that it becomes true ?

Let $p:K\to B$ be an acyclic fibration and a proper map. Assume $p:K\to B$ is "nice" enough (thus, e.g. all notions of "fibration" are equivalent for maps so nice). Let $f:K\to K$ be an endomorphism fixing $p$, i.e. $p\circ f=p$. Does there necessarily exist a section $s:B\to K$ fixed by $f:K\to K$, i.e. $s=f\circ s$.

The counterexample is as follows, adapted from Is it possible to continuously select a probability distribution over fixed points in Brouwer's fixed point theorem?. The map $p:K\to B$ is the projection $[0,1]\times [0,1]\to [0,1]$.

The endomorphism $f:[0,1]\times [0,1]\to[0,1]\times [0,1]$ is the family of functions $f_t:[0,1]\rightarrow[0,1]$ ($t\in[0,1]$) such that $f_0(y)=\frac{1}{3}$; $f_{\frac{1}{2}}(y)=\frac{1}{3}$ for $y\leq\frac{1}{3}$, $f_{\frac{1}{2}}(y)=\frac{2}{3}$ for $y\geq\frac{2}{3}$, and linearly interpolates between those for $\frac{1}{3}<y<\frac{2}{3}$; $f_1(y)=\frac{2}{3}$; and $f_t$ linearly interpolates between $f_0$ and $f_\frac{1}{2}$ for $0<t<\frac{1}{2}$ and between $f_\frac{1}{2}$ and $f_1$ for $\frac{1}{2}<t<1$.

$1/3$, resp. $2/3$ is the only fixed point of $f_t$ for $t<1/2$, resp. $t>1/2$. For $t=1/2$ the fixed points of $f_{1/2}$ form the interval [1/3,2/3]$.

Thus, no section exists.

Is there a version for the Brouwer fixed point theorem for maps rather than spaces ? In other words, for a family of endomorphisms, can the fixed point be chosen continuously, under some assumptions ?

One way to make this question precise is as follows.

A standard formulation of the Brouwer fixed point theorem says that there is a fixed point of an endomorphism of a contractible compact space $K$, whenever the space is sufficiently "nice". In terms of maps, it means that the map $K\to \{pt\}$ is an acyclic fibration and a proper map.

Is there a version of the Brouwer fixed point theorem for "endomorphisms" of maps which are both acyclic fibration and a proper map ?

Below I give a naive version is false due to the counterexample below. Can one make stronger assumptions such that it becomes true, or a weaker conclusion ? For example, as suggested by Tom Goodwillie in comments,

Let $F\subset K$ be the set of fixed points. I wonder if there is a chance if showing that the projection $F\to B$ has a right inverse up to homotopy (under some broad hypothesis). – Tom Goodwillie

The following is a false naive version.

Let $p:K\to B$ be an acyclic fibration and a proper map. Assume $p:K\to B$ is "nice" enough (thus, e.g. all notions of "fibration" are equivalent for maps so nice). Let $f:K\to K$ be an endomorphism fixing $p$, i.e. $p\circ f=p$. Does there necessarily exist a section $s:B\to K$ fixed by $f:K\to K$, i.e. $s=f\circ s$.

The counterexample is as follows, adapted from Is it possible to continuously select a probability distribution over fixed points in Brouwer's fixed point theorem?. The map $p:K\to B$ is the projection $[0,1]\times [0,1]\to [0,1]$.

The endomorphism $f:[0,1]\times [0,1]\to[0,1]\times [0,1]$ is the family of functions $f_t:[0,1]\rightarrow[0,1]$ ($t\in[0,1]$) such that $f_0(y)=\frac{1}{3}$; $f_{\frac{1}{2}}(y)=\frac{1}{3}$ for $y\leq\frac{1}{3}$, $f_{\frac{1}{2}}(y)=\frac{2}{3}$ for $y\geq\frac{2}{3}$, and linearly interpolates between those for $\frac{1}{3}<y<\frac{2}{3}$; $f_1(y)=\frac{2}{3}$; and $f_t$ linearly interpolates between $f_0$ and $f_\frac{1}{2}$ for $0<t<\frac{1}{2}$ and between $f_\frac{1}{2}$ and $f_1$ for $\frac{1}{2}<t<1$.

$1/3$, resp. $2/3$ is the only fixed point of $f_t$ for $t<1/2$, resp. $t>1/2$. For $t=1/2$ the fixed points of $f_{1/2}$ form the interval $[1/3,2/3]$.

Thus, no section exists.

Is there a version for the Brouwer fixed point theorem for maps rather than spaces ? In other words, for a family of endomorphisms, can the fixed point be chosen continuously, under some assumptions ?

One way to make this question precise is as follows.

A standard formulation of the Brouwer fixed point theorem says that there is a fixed point of an endomorphism of a contractible compact space $K$, whenever the space is sufficiently "nice". In terms of maps, it means that the map $K\to \{pt\}$ is an acyclic fibration and a proper map.

Is there a version of the Brouwer fixed point theorem for "endomorphisms" of maps which are both acyclic fibration and a proper map ?

For example, the following naive version is false due to the counterexample below. Can one make stronger assumptions such that it becomes true ?

Let $p:K\to B$ be an acyclic fibration and a proper map. Assume $p:K\to B$ is "nice" enough (thus, e.g. all notions of "fibration" are equivalent for maps so nice). Let $f:K\to K$ be an endomorphism fixing $p$, i.e. $p\circ f=p$. Does there necessarily exist a section $s:B\to K$ fixed by $f:K\to K$, i.e. $s=f\circ s$.

The counterexample is as follows, adapted from Is it possible to continuously select a probability distribution over fixed points in Brouwer's fixed point theorem?. The map $p:K\to B$ is the projection $[0,1]\times [0,1]\to [0,1]$.

The endomorphism $f:[0,1]\times [0,1]\to[0,1]\times [0,1]$ is the family of functions $f_t:[0,1]\rightarrow[0,1]$ ($t\in[0,1]$) such that $f_0(y)=\frac{1}{3}$; $f_{\frac{1}{2}}(y)=\frac{1}{3}$ for $y\leq\frac{1}{3}$, $f_{\frac{1}{2}}(y)=\frac{2}{3}$ for $y\geq\frac{2}{3}$, and linearly interpolates between those for $\frac{1}{3}<y<\frac{2}{3}$; $f_1(y)=\frac{2}{3}$; and $f_t$ linearly interpolates between $f_0$ and $f_\frac{1}{2}$ for $0<t<\frac{1}{2}$ and between $f_\frac{1}{2}$ and $f_1$ for $\frac{1}{2}<t<1$.

$1/3$, resp. $2/3$ is the only fixed point of $f_t$ for $t<1/2$, resp. $t>1/2$. For $t=1/2$ $f_{1/2}$ has two fixed points $1/3$ and $2/3$.

Thus, no section exists.

Is there a version for the Brouwer fixed point theorem for maps rather than spaces ? In other words, for a family of endomorphisms, can the fixed point be chosen continuously, under some assumptions ?

One way to make this question precise is as follows.

A standard formulation of the Brouwer fixed point theorem says that there is a fixed point of an endomorphism of a contractible compact space $K$, whenever the space is sufficiently "nice". In terms of maps, it means that the map $K\to \{pt\}$ is an acyclic fibration and a proper map.

Is there a version of the Brouwer fixed point theorem for "endomorphisms" of maps which are both acyclic fibration and a proper map ?

For example, the following naive version is false due to the counterexample below. Can one make stronger assumptions such that it becomes true ?

Let $p:K\to B$ be an acyclic fibration and a proper map. Assume $p:K\to B$ is "nice" enough (thus, e.g. all notions of "fibration" are equivalent for maps so nice). Let $f:K\to K$ be an endomorphism fixing $p$, i.e. $p\circ f=p$. Does there necessarily exist a section $s:B\to K$ fixed by $f:K\to K$, i.e. $s=f\circ s$.

The counterexample is as follows, adapted from Is it possible to continuously select a probability distribution over fixed points in Brouwer's fixed point theorem?. The map $p:K\to B$ is the projection $[0,1]\times [0,1]\to [0,1]$.

The endomorphism $f:[0,1]\times [0,1]\to[0,1]\times [0,1]$ is the family of functions $f_t:[0,1]\rightarrow[0,1]$ ($t\in[0,1]$) such that $f_0(y)=\frac{1}{3}$; $f_{\frac{1}{2}}(y)=\frac{1}{3}$ for $y\leq\frac{1}{3}$, $f_{\frac{1}{2}}(y)=\frac{2}{3}$ for $y\geq\frac{2}{3}$, and linearly interpolates between those for $\frac{1}{3}<y<\frac{2}{3}$; $f_1(y)=\frac{2}{3}$; and $f_t$ linearly interpolates between $f_0$ and $f_\frac{1}{2}$ for $0<t<\frac{1}{2}$ and between $f_\frac{1}{2}$ and $f_1$ for $\frac{1}{2}<t<1$.

$1/3$, resp. $2/3$ is the only fixed point of $f_t$ for $t<1/2$, resp. $t>1/2$. For $t=1/2$ the fixed points of $f_{1/2}$ form the interval [1/3,2/3]$.

Thus, no section exists.