For which continua $X$ does $X^\mathbb{N}$ have the fixed-point property?

Question

A topological space $X$ has the fixed-point property if any continuous map $f : X \to X$ has a fixed point (i.e., an $x$ such that $f(x) = x$). It's well known that certain topological spaces, such as $[0,1]^n$ by Brouwer's fixed-point theorem, have the fixed-point property. It's also known that even among relatively tame spaces, the property can be poorly behaved or at least subtle under passing to product spaces. For examples, see this or this MathOverflow question.

On the other hand, the operation of passing from the space $X$ to the space $X^\mathbb{N}$ sometimes has the effect of uniformizing the space in question. In the context of continua (i.e., non-empty compact connected metric spaces) in particular, it's well known that $X^\mathbb{N}$ is homeomorphic to the Hilbert cube, $[0,1]^\mathbb{N}$, for any dendrite $X$. The Hilbert cube is even more regular than the interval itself, in that the Hilbert cube is homogeneous despite the fact that the interval is not.

This leads me to the following question:

Question. For which continua $X$ does $X^\mathbb{N}$ have the fixed-point property?

One thing to note is that the fixed-point property for $X^\mathbb{N}$ easily implies the fixed-point property for $X$ (and $X^n$ for all $n$), so this is a strictly stronger condition than the fixed-point property itself.

In looking through the literature, I have only been able to find this possibly relevant result, due to West: If $X$ is a topological space with more than one point such that $X \times [0,1]^\mathbb{N}$ is homeomorphic to $[0,1]^\mathbb{N}$, then $X^\mathbb{N}$ is homeomorphic to $[0,1]^\mathbb{N}$.

Answer 1

Let $X$ be a compact metric space. Then $X^\omega$ has the fixed point property if and only if $X^N$ has the fixed point property for all $N$. We observe that every retract of a space with the fixed point property has the fixed point property, so if $X^\omega$ has the fixed point property, then so does each $X^N$ (the asker has already observed this fact). Let $0\in X$. Let $X^N_e$ be the collection of all tuples $(x_n)_{n\in\omega}$ where $x_n=0$ for $n\geq N$ ($X_e^N$ is a homeomorphic copy of $X^N$). Let $H_N:X^\omega\rightarrow X_e^N$ be the canonical retraction obtained by setting the $n$-th coordinate to $0$ whenever $n\geq N$.

Now suppose that $f:X^\omega\rightarrow X^\omega$ and $X^N$ has the fixed point property for each natural number $N$. Then for all natural numbers $N$, the mapping $H_N\circ f$ has some fixed point $(x_{n,N})_{n\in\omega}$.

Therefore, there is $(x_n)_{n\in\omega}$ and infinite subset $A\subseteq\omega$ such that $\lim_{N\rightarrow\infty,N\in A}(x_{n,N})_{n\in\omega}\rightarrow(x_n)_{n\in\omega}.$ By continuity, we know that

$$f(x_n)_{n\in\omega}=\lim_{N\rightarrow\infty,N\in A}f(x_{n,N})_{n\in\omega}=\lim_{N\rightarrow\infty,N\in A}H_N\circ f(x_{n,N})_{n\in\omega}$$ $$=\lim_{N\rightarrow\infty,N\in A}(x_{n,N})_{n\in\omega}=(x_n)_{n\in\omega}.$$