Proof of Lemma 1.

Without loss of generality consider the network ①. There is a chain i1,i2,i3,i4 . We will show that if i2≿i1i3 then i2≿i4i3 . To do so, let us assume that i2≿i1i3 .^[11] This assumption necessitates that in the network ②, which is the viewpoint of i1 that contains both agents i2 and i3 , i2 is better informed than i3 , where the term ‘better informed’ refers to the fact that i2 has more total ex-post information than i3 does.

Now let us modify the network ② by eliminating the link gˉi3i4=1 , which results in the network ③. Observe that by removing gˉi3i4=1 from the network ②, both i2 and i3 lose information from i4 . However, i3 loses more information from i4 than i2 does because i3 is closer to i4 than i2 . This fact and the fact that i2 is better informed than i3 in ② imply that i2 is also better informed than i3 in ③.

Next, let us modify the network ③ by adding the link gi2i1=1 so that the network ③ becomes the network ④. Observe that in ④ the agent i2 is closer to i1 than i3 is. This observation together with the fact that in ③ i2 is better informed than i3 and that the network ④ is simply ④ = ③ + i1i2 lead to the conclusion that in ④ i2 is better informed than i3 . Finally, observe that ④ is nothing else but the viewpoint of i3 that contains i2 and i1 . This fact and the aforementioned fact that in ④ i2 is better informed than i3 allow us to conclude that i2≿i4i3 .   □

Proof of Lemma 2.

Let us consider a chain i1,i2,...,in−1,in . Without loss of generality let us assume that i2≿i1in−1 . Consequently our goal is to show that i2≿inin−1 . Before so doing, we remark that in the model of Charoensook (2015) and any model that assumes no information decay any agent i prefers agent x to y if and only if ci,x≤ci,y . Onwards, we use this fact to complete this proof.

Now since it is assumed that i2≿i1in−1 we know that ci1,i2≤ci1,in−1 . Due to the fact that C={ci,j:i,j∈N,i≠j} satisfies UPR, ci1,i2≤ci1,in−1 necessitates that cin,i2≤cin,in−1 , which in turn necessitates that i2≿i1in−1 .   □

Proof of Lemma 3.

[If a minimally connected network has no inward-pointing chain with more than three agents, then this network is either Bi or branching]

The absence of inward-pointing chain with more than three agents results in the fact that there are three types of four-agent chain in a minimally connected network. The first type sequentially consists of links {i1i2,i2i3,i3i4} . The second type sequentially consists of links {i2i1,i2i3,i3i4} . The third type sequentially consists of links {i1i2,i3i2,i3i4} . Note that the first and the second types are such that there is no agent who receives more than one link, while the third case is such that there is exactly one agent who receives more than one link. Onwards we split our proof into two subsections. In the first subsection, we show that the absence of the third type leads to a network that is branching. On the contrary, in the second subsection we show that the presence of the third type leads to a Bi network.

Let us proceed to the first subsection. First, note that since the absence of the third type of four-agent chain is assumed, the network consists of chains that are of either first type or second type or both. We claim that in any case this network is a branching network. Assuming, without loss of generality, that a chain of the first type exists, recall that the sequence of link in this chain is {i1i2,i2i3,i3i4} .^[12] We first show that there is at least one agent who receives no link. Suppose not. Then it is the case that i1 receives a link from another agent j1 , who receives a link from another agent j2 . This analogy repeats infinitely, causing the network to consist of an infinite number of agents. A contradiction. Hence, there is at least one agent who receives no link. Next, we show that there is exactly one such agent. Let this agent be i∗ . The fact that i∗ receives no link necessitates that i∗ establishes a link with all his adjacent agents. Hence, each of his adjacent agents receives exactly one link. The fact that each of these neighbors receives exactly one link from i∗ further necessitates that he establishes links with all his adjacent agents who are not i∗ . Thus, beside i∗ every other agent receives exactly one link and establishes at least one link unless he is a terminal agent. We conclude, therefore, that this network is a branching network.

Next, we proceed to the second subsection, which is to show that if the third type of chain – {i1i2,i3i2,i3i4} – is present then this network is a Bi network. Indeed, we can show that this network is Bi2 network. To do so we let j be an agent who is a adjacent agent of one of these four agents – i1,i2,i3 and i4 . Clearly j receives a link from one of these four agents. Otherwise, it is straightforward to show that an inward-pointing chain with more than three agents exist. For example if j is a adjacent agent of i1 and j establishes a link with i1 , then the chain ji1,i1i2,i3i2 is inward pointing. Similarly, any adjacent agent of j that is not one of these four agents also receives a link from j. By repeating this analogy, we have a path from any adjacent agent of i2 to a terminal agent in this network. Put differently, any agent in this network can be reached through a path from an agent that is an adjacent agent of i2 . It follows that a subset of the set of all adjacent agents of i2 forms a contrabasis of this network. Finally, observe that since there is a path from any adjacent agent of i2 to a terminal agent in this network, i2 is the only agents who receives more than one link and every other agent receives exactly one link. We conclude, therefore, that this network is Bi2 network.

If a minimally connected network is either a branching or Bi network, there exists no inward-pointing chain in this component.

To do so, we prove that if there exists an inward-pointing chain with more than three agents in a minimally connected network, then the network is neither a branching or Bi network. We divide our proof into two steps: (i) if an inward-pointing chain with more than three agents exists, then the network has at least one agent i who receives more than one link, which necessitates that the network is not a branching and (ii) this agent i is not an i− point contrabasis of this network, which necessitates that the network is not a Bi network.

Let us prove the first step: if an inward-pointing chain with more than three agents exists, then the network has at least one agent i who receives more than one link. Consider an inward-pointing chain with more than three agents. Let this chain be between i and j. Let i accesses i’ and j accesses j’ in this chain. Next, to prove by contradiction let us suppose that there is no agent who receives more than one link. Then since i accesses i’, we know that i’ also accesses his adjacent agent in this chain. By repeating this analogy we know that j’ access j. A contradiction. It follows that this inward-pointing chain with more than three agents contains at least one agent who receives more than one link. Let this agent be i∗ .

We now prove the second step: i∗ is not an i∗ -point contrabasis of this network. First recall from the previous paragraph that i∗ resides in an inward-pointing chain with more than three agents, and this chain is between i and j where i∗≠i,j . Next, consider a chain between i∗ and j. To prove by contradiction let us suppose that i∗ is a point contrabasis and the network is Bi∗ . The assumption that i∗ is a point contrabasis necessitates that there is a path from a adjacent agent of i∗ to j. Let this adjacent agent of i∗ be k. Since a path from k to j exists, we know that k accesses one of his adjacent agents, and this agent access another agent. By repeating this analogy we know that j’ accesses j. A contradiction to the assumption that j accesses j’ in this inward-pointing chain.   □