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Oct 29, 2012 at 20:35 comment added KP Hart The valency of the graph is $2^{\mathfrak{c}}$, rather than $\mathfrak{c}$; all points of the remainder have have $2^{\mathfrak{c}}$ preimages under non-trivial finite-to-one maps from $\mathbb{N}$ to itself (those where the fibers have arbitrarily large finite cardinality).
Oct 25, 2012 at 18:57 comment added YCor I maybe miss something, as I don't see how this question can be open: if you consider the oriented labeled graph with the Stone-Cech boundary as set of vertices, and oriented edge $(x,y)$ labeled by $f:I\to J$ whenever $f(x)=y$, and consider the underlying unoriented graph, then its valency is at most continuum, and therefore its components have cardinality at most continuum. So given $x$ you can find $y$ in another component.
May 12, 2010 at 8:10 history answered KP Hart CC BY-SA 2.5