For $S$ a set, let $\beta_{\bf2}(S)$ be a compact, totally disconnected space containing $S$ where $S$ in the subspace topology is discrete and $S$ is a dense subspace, and $\beta_{\bf2}(S)$ has the property that, for any compact, totally disconnected space $U$ and any function $f:S\to U$, there is a continuous map from $\beta_{\bf2}(S)$ to $U$ extending $f$.

Consider $\beta_{\bf2}(S)$ where $S=T\times\{0,1\}$ for some set $T$.

I imagine it is a disjoint union of two spaces homeomorphic to $\beta_{\bf2}(T)$ where each space is a clopen subset of the whole. If so, let us call the entire space $\beta_{\bf2}(T)\times\{0,1\}$, with the obvious meaning.

Let $U$ be a compact, totally disconnected space.

Let $f:T\times\{0,1\}\to U$ be a function such that, for all $t\in T$, $f(t,0)\ne f(t,1)$.

I assume $f$ extends to a continuous map from $\beta_{\bf2}(T)\times\{0,1\}$ to $U$.

Can we say that for all $b\in\beta_{\bf2}(T)\times\{0,1\}$, $f(b,0)\ne f(b,1)$?

What if we fix $u_0\in U$ and say that, for all $t\in T$, it is not the case that $f(t,0)=u_0=f(t,1)$.

Can there be $b\in\beta_{\bf2}(T)\times\{0,1\}$ such that $f(b,0)=u_0=f(b,1)$?

For $S$ a set, let $\beta_{\bf2}(S)$ be a compact, totally disconnected space containing $S$ where $S$ in the subspace topology is discrete and $S$ is a dense subspace, and $\beta_{\bf2}(S)$ has the property that, for any compact, totally disconnected space $U$ and any function $f:S\to U$, there is a continuous map from $\beta_{\bf2}(S)$ to $U$ extending $f$.

Consider $\beta_{\bf2}(S)$ where $S=T\times\{0,1\}$ for some set $T$.

I imagine it is a disjoint union of two spaces homeomorphic to $\beta_{\bf2}(T)$ where each space is a clopen subset of the whole. If so, let us call the entire space $\beta_{\bf2}(T)\times\{0,1\}$, with the obvious meaning.

Let $U$ be a compact, totally disconnected space.

Let $f:T\times\{0,1\}\to U$ be a function such that, for all $t\in T$, $f(t,0)\ne f(t,1)$.

I assume $f$ extends to a continuous map from $\beta_{\bf2}(T)\times\{0,1\}$ to $U$. Let us also call the extension $f$.

Can we say that for all $b\in\beta_{\bf2}(T)$, $f(b,0)\ne f(b,1)$?

What if we fix $u_0\in U$ and say that, for all $t\in T$, it is not the case that $f(t,0)=u_0=f(t,1)$.

Can there be $b\in\beta_{\bf2}(T)$ such that $f(b,0)=u_0=f(b,1)$?

For $S$ a set, let $\beta_{\bf2}(S)$ be a compact, totally disconnected space containing $S$ where $S$ in the subspace topology is discrete and $S$ is a dense subspace, and $\beta_{\bf2}(S)$ has the property that, for any compact, totally disconnected space $U$ and any function $f:S\to U$, there is a continuous map from $\beta_{\bf2}(S)$ to $U$ extending $f$.

Consider $\beta_{\bf2}(S)$ where $S=T\times\{0,1\}$ for some set $T$.

I imagine it is a disjoint union of two spaces homeomorphic to $\beta_{\bf2}(T)$ where each space is a clopen subset of the whole. If so, let us call the entire space $\beta_{\bf2}(T)\times\{0,1\}$, with the obvious meaning.

Let $U$ be a compact, totally disconnected space.

Let $f:T\times\{0,1\}\to U$ be a function such that, for all $t\in T$, $f(t,0)\ne f(t,1)$.

I assume $f$ extends to a continuous map from $\beta_{\bf2}(T)\times\{0,1\}$ to $U$.

Can we say that for all $b\in\beta_{\bf2}(T)\times\{0,1\}$, $f(b,0)\ne f(b,1)$?

For $S$ a set, let $\beta_{\bf2}(S)$ be a compact, totally disconnected space containing $S$ where $S$ in the subspace topology is discrete and $S$ is a dense subspace, and $\beta_{\bf2}(S)$ has the property that, for any compact, totally disconnected space $U$ and any function $f:S\to U$, there is a continuous map from $\beta_{\bf2}(S)$ to $U$ extending $f$.

Consider $\beta_{\bf2}(S)$ where $S=T\times\{0,1\}$ for some set $T$.

I imagine it is a disjoint union of two spaces homeomorphic to $\beta_{\bf2}(T)$ where each space is a clopen subset of the whole. If so, let us call the entire space $\beta_{\bf2}(T)\times\{0,1\}$, with the obvious meaning.

Let $U$ be a compact, totally disconnected space.

Let $f:T\times\{0,1\}\to U$ be a function such that, for all $t\in T$, $f(t,0)\ne f(t,1)$.

I assume $f$ extends to a continuous map from $\beta_{\bf2}(T)\times\{0,1\}$ to $U$.

Can we say that for all $b\in\beta_{\bf2}(T)\times\{0,1\}$, $f(b,0)\ne f(b,1)$?

What if we fix $u_0\in U$ and say that, for all $t\in T$, it is not the case that $f(t,0)=u_0=f(t,1)$.

Can there be $b\in\beta_{\bf2}(T)\times\{0,1\}$ such that $f(b,0)=u_0=f(b,1)$?