$\omega$-pure subgraph in $\mathbf{Grph}$ Here
on the page 7:
I do not follow how $B$ looks like. It is written there:

$B$ is a strong subgraph of $A$ over all nodes distinct from $x_k$ for $k\geq 1$".

So $B$ is only $x_0$ as we have removed all $x_k$ for $k\geq 1$ ??
So, my question is: what is $B$, why it is strong and why $B$ is not in $\cal A$ ?
I just do not follow that example.

 A: The graphs $A$ and $B$ have many other nodes beside the $x_k$.
First, to clarify their construction:
(1) Start with the disjoint union of the following graphs:

*

*for each $n \in \mathbb{N}$, a directed path of length n:
$v_{n,0} \to v_{n,1} \to ... \to v_{n,n}$


*one infinite directed path:
$x_0 \to x_1 \to \cdots$
(2) Then identify all nodes $v_{n,0}$ $(n>0)$ with $x_0$.
This is the graph $A$. It has a starting node $x_0$ to which all those pathes
are attached.
The graph $B$ is obtained by throwing away the $x_k$ for $k>1$.
What remains is the starting node $x_0$ to which all the finite pathes
are attached.
(a) $B$ is not in ${\cal A}$ because it has no infinite directed path.
(b) $B$ is a full subgraph of $A$ and therefore the inclusion of $B \hookrightarrow A$
is a strong monomorphism i.e. it has the right lifting property w.r.t.
all epimorphisms. The proof is the same as for sets.
It is perhaps not obvious how $\omega$-purity of $B \hookrightarrow A$ follows
from the last few lines in the example, so I will fill in the details:
Remember that $\omega$-presentable means finitely presentable
and for graphs finitely presentable is the same as finite.
That $B \hookrightarrow A$ is $\omega$-pure then means that for any given commuting square
$$
\matrix{
 &   &       u     &   & \\
 & X & \rightarrow & Y & \\
f& \downarrow &  & \downarrow & g \\
 & B & \hookrightarrow & A &
}
$$
where $X$ and $Y$ are finite graphs, the map $f$ must factor through $u$
via some map from Y to B.
Now consider the image $g(Y)$ of $g$. This is a finite subgraph of $A$.
The pullback of $B \hookrightarrow A$ and $g(Y) \hookrightarrow A$ is
just the intersection $D = B \cap g(Y)$ , which is also finite.
So we end up with
$$
\matrix{
 &   &       u     &   & \\
 & X & \rightarrow & Y & \\
 & \downarrow &  & \downarrow & g \\
 & D & \hookrightarrow & g(Y) & \\
 & \downarrow &  & \downarrow &  \\
 & B & \hookrightarrow & A &
}
$$
where the outer rectangle is the original square from above.
Now it is enough to show that the inclusion of $D \hookrightarrow B$ factors through
the inclusion $D \hookrightarrow g(Y)$ via some map from $g(Y)$ to $B$. Then $f$ will
factor through $u$ via the composition $Y \to g(Y) \to B$.
So the whole condition of omega-purity of $B \hookrightarrow A$ is reduced to the
condition: every finite subgraph $V \subseteq A$ has a map to $B$ which is the
identity on $B$. This is what is shown at the end of the example.
Now, if $V$ is a subgraph of $A$, then a node of $V \setminus B$ must lie
on the infinite path of $A$. If $V$ is finite, then $V \setminus B$
is finite and hence we can find a natural number $n$
such that the nodes of $V \setminus B$ are among the nodes $x_0, \ldots , x_n$.
But we also have a finite path $x_0, x'_1, \ldots , x'_n$ in $B$.
So, for $x \in V \setminus B$  we have $x = x_k$ for some $k\leq n$ and
we can set $h(x_k) = x'_k$. For $x \in B$, we can set $h(x) = x$.
This gives the desired map $h: V \to B$
