Directed Graph with restriction such that each node can only have at most one incoming edge My intuition is that a directed graph such that each node can have at most one "incoming edge" (edge which is pointed to the node), can only have at most 1 cycle.
I've tried constructing a few by hand:

Is my intuition correct? Is there a formal proof for this? In this specific setup, is there a method for determining where the cycle is, and removing one edge so the graph becomes a DAG (directed acyclic graph)?

Expanding on the comment I made earlier:
Let $V$ be the set of vertices, and define the partial function
$f \colon V \to V,$ where for each $x \in V,$ $f(x)$ is the unique
vertex with an edge going to $x,$ if there is a such a vertex, and
otherwise $f(x)$ is undefined.
If $V$ is weakly connected, then it is clear from the
definition (see, e.g., Connected Digraph -- from Wolfram
MathWorld) that no
non-empty proper subset of $V$ is both $f$-closed and
$f^{-1}$-closed.
That is, if $S \subseteq V,$ $S \ne \emptyset,$ $f(S) \subseteq S$
and $f^{-1}(S) \subseteq S,$ then $S = V.$
For each $x \in V,$ let $f^*(x)$ be the intersection of all
$f$-closed subsets of $V$ containing $x.$ Then
$$
f^*(x) = \{f^n(x) : n \in \mathbb{N} \text{ and }
f^n(x) \text{ is defined}\},
$$
where $f^0(x) = x$ and $f^{n+1}(x) = f(f^n(x))$ for all
$n \in \mathbb{N}$ such that $f^n(x)$ is defined and $f(f^n(x))$ is
defined.
If $K$ is a cycle, then $K = f^*(x)$ for all $x \in K.$
For all $A \subseteq V,$ define
$$
\breve{f}^*(A) = \{x \in V : f^*(x) \cap A \ne \emptyset\} =
\{x \in V : f^n(x) \in A \text { for some } n \in \mathbb{N} \}.
$$
Clearly $\breve{f}^*(A)$ is $f^{-1}$-closed. (It is the intersection
of all $f^{-1}$-closed supersets of $A,$ but we don't need this.)
If $A$ is $f$-closed, in particular if $A$
is a cycle, then $\breve{f}^*(A)$ is also $f$-closed.
(Proof: consider separately the cases $n = 0$ and $n > 0.$)
Therefore, if $V$ is weakly connected and $K$ is a cycle then
$V = \breve{f}^*(K).$
Suppose $V$ is weakly connected, and $K$ and $L$ are cycles.  Take
any $x \in K.$ Then
$$
x \in V = \breve{f}^*(L) =
\{x \in V : f^*(x) \cap L \ne \emptyset\}.
$$
But $f^*(x) = K,$ therefore $K \cap L \ne \emptyset.$
Taking any $y \in K \cap L,$ we have $K = f^*(y) = L.$
That is, $V$ contains at most one cycle. $\ \square$
If $V$ is weakly connected, and has a cycle, $K,$ then, starting
from any vertex $x \in V,$ we can find an element $f^n(x) \in K$ for
some natural number $n$; and then, by deleting any edge from $K,$
we can turn $V$ into a DAG.
