Conjecture about graph with points and arrows with weights

Question

Postulates

1. Each point has arrow(s), and each arrow is pointing to another point.
2. An arrow of a point cannot point to itself.
3. Arrows can only point to any direction on the right. That means $\uparrow$, $\downarrow$ are not allowed, and of course $\leftarrow$ is not allowed.
4. Each arrow has a number $x$ on it with $0<x<1$ and $x\in\mathbb{R}$.
5. Of a point, say A, arrows pointing out to the right must have their numbers summing up to 1, only if the arrows exist. (That means if a point does not have arrows pointing to the right, the condition does not have to be satisfied) Same for arrows pointing to A.

Conjecture

Suppose there are a few points and arrows. Prove that if there are some points each with some arrows pointing to the right, there must be the same case but being pointed from the left, in order to satisfy the conditions listed above. (At least this is the pattern I can see.)

For example, if there are 2 points each with 3 arrows each pointing to another point, there must be two points each being pointed with 3 arrows.

I think it is related to Graph Theory, but I have no idea of proving it, or better say I am not familiar with Graph Theory.

Evidences

Example 1: There exists $A\xrightarrow{1} B$. Since there is 1 point with 1 arrow pointing to another point, there must be and there is 1 point being pointed with 1 arrow.

Example 2: point A and B both have two arrows pointing to C and D. Hence, both point C and D are being pointed by two arrows from A and B.

Example 3: point A and C both have two arrows pointing out, so two points D and F are being pointed by two arrows. point B has three arrows pointing out, so a point E is being pointed by three arrows.

This graph also satisfies the conditions above:

1. For point A: $0.8+0.2=1$
2. For point B: $0.2+0.3+0.5=1$
3. For point C: $0.3+0.7=1$
4. For point D: $0.8+0.2=1$
5. For point E: $0.2+0.3+0.5=1$
6. For point F: $0.3+0.7=1$

These are just simple cases. The conjecture will not be obvious or maybe even false in more complex graphs.

Answer 1

This is not true. For example, consider a network with four nodes $A,B,C,D$ on the left and four nodes $E,F,G,H$ on the right. Give $A$ two arcs of weight $0.5$ to $E,F$. Give $D$ two arcs of weight $0.5$ to $G,H$. Give each of $B,C$ four arcs of weight $0.25$ to $E,F,G,H$. Then every node on the left has $1$ unit going out, and every node on the right has $0.5+0.25+0.25=1$ unit coming in. But all nodes on the left have $2$ or $4$ arcs going out, and all nodes on the right have $3$ coming in.