Inducing maps between Martin boundaries This is a reworking of a question I asked on math.se.
Given two countable discrete metric spaces $X_{1}$ and $X_{2}$, each equipped with a (irreducible and transient)* random walk given by transition functions $p_{1}$, and $p_{2}$ what conditions must be satisfied so that a function $f:X_{1}\rightarrow X_{2}$ induces a continuous map from the Martin boundary on $X_{1}$ to the Martin boundary on $X_{2}$.
The following question is related, but doesn't seem to fit the bill:How does a quasi-isometry affect Poisson or Martin boundaries?
Looking through the literature I don't see much (any) attention given to relationships between random walks on different spaces, but surely such relationships must be well known. An answer to my question above or else a reference would greatly appreciated.
*Given a random walk with transition function $p$ on a countable metric space $X$ we define $p_{0}(x,y)=\delta_{xy},\,p_{1}(x,y)=p(x,y)$ and for $n\geq 2$ we define
$$p_{n}(x,y)=\sum_{z\in X}p_{n-
1}(x,z)p_{1}(z,y)$$
Then we define the Green's function $g(x,y)$ by
$$g(x,y)=\sum_{n=0}^{\infty}p_{n}(x,y)$$
We call the random walk transient if $g(x,y)<\infty$ fo reach $x,y\in X$. We call the random walk irreducible if for all $x,y\in X$ there is some $n$ such that $p_{n}(x,y)>0$.
 A: First, although I don't have a precise statement, the Martin boundary seems more contravariant than covariant, in the sense that any boundary $\partial X$ of $X$ that satisfies that $X$ is dense in $X\cup \partial X$ and that the Martin kernels $K(x,y)$ extend continuously on $X\times X\cup \partial X$ is a quotient of the whole Martin boundary. So one would maybe rather expect a map from the Martin boundary of $X_2$ to the Martin boundary of $X_1$. However, it does not seem that there are natural conditions to ensure the existence of such a map, as I try to explain in the following discussion.
One very strong but natural condition would be that $f$ preserves harmonicity, in the following sense. Assuming that $h_2$ is a $p_2$-harmonic function on $X_2$, then $h_1=h_2\circ f$ is $p_1$-harmonic on $X_1$.
Now consider the discrete Heisenberg group $G=H_3(\mathbb{Z})=\langle a,b,c|c=[a,b],[a,c]=e,[b,c]=e\rangle$, whose abelianization $G/H$ is isomorphic to $\mathbb{Z}^2$, where $H=[G,G]$. Let $\pi$ be the canonical projection of $G$ onto $G/H$ and choose $X_1=G$, $X_2=G/H$ and $f=\pi$. Also consider a non-centered finitely supported probability measure $p_1$ on $G$ and define $p_2(gH)=\sum_{x\in H}p_1(gx)$. So in other words $p_2=f_*p_1$. Here, $f$ preserves harmonicity. Indeed, let $\phi$ be $p_2$-harmonic, then
$$\sum_{y\in G}p_1(x^{-1}y)\phi \circ f(y)=\sum_{yH}\sum_{h\in H}p_1(x^{-1}yh)\phi(yH)=\sum_{\tilde{y}\in G/H}p_2(\tilde{x}^{-1}\tilde{y})\phi(\tilde{y})=\phi(\tilde{x})=\phi\circ f(x).$$
Actually, althouth we do not need this, note that in this particular example, since $G$ is nilpotent, any $p_1$-harmonic function is constant on the left cosets of $H$ according to a landmark result of Margulis. Hence, any $p_1$-harmonic function defines a $p_2$-harmonic function.
Conjecturally, the Martin boundary of $(X_1,p_1)$ is homeomoprhic to a disk $\mathbb{D}= \{z\in \mathbb{C}, |z|\leq 1\}$ and the minimal Martin boundary is a circle and coincides with $\partial \mathbb{D}=\{z\in \mathbb{C},|z|=1\}$. On the other hand, the Martin boundary of $G/H$ is the Martin boundary of a non-centered probability measure on an abelian group and so it is homeomorphic to the circle. This conjecture comes from the corresponding result for the Brownian motion on the real Heisenberg group $\mathbb{H}^3(\mathbb{R})$, see here.
You see that there is no canonical continuous map from the Martin boundary of $X_1$ to the Martin boundary of $X_2$, although the function $f$ is very nice wrt $p_1$ and $p_2$. Also, even if you can abstractly consider the Martin boundary of $X_2$ as a subset of the Martin boundary of $X_1$, you don't have a very canonical way of defining a map $\partial X_2\to \partial X_1$. Typically, to define such a map you would want that a sequence $y_n\in X_2$ that converges to a point in the Martin boundary of $X_2$ satisfies that any preimage $x_n\in f^{-1}(y_n)$ converges to a uniquely defined point in the Martin boundary of $X_1$. However, in our example, you probably don't have this property (see Section 3 of the above cited paper, where there is a precise description of the topology of the Martin boundary of $X_1$).
