Unanimity and local incentive compatibility in sparsely connected domains

Abstract

This paper studies the implications of imposing unanimity and local incentive compatibility on a deterministic social choice function. In an environment with strict ordinal preferences over a finite set of alternatives, we find that tops-onlyness and full incentive compatibility necessarily follow from unanimity and local incentive compatibility in sparsely connected domains. Furthermore, we identify a property of preference domains that completely characterizes dictatorial domains within sparsely connected domains.

Appendix

Proof of Proposition 1

The first step of our proof is to show that when the top alternatives are fixed, the outcome of f is invariant to adjacent manipulations in preferences (Lemma 4). To do so, we introduce a property called swap monotonicity and a result—both from Mishra (2016)—that any G-LOBIC SCF satisfies swap monotonicity (Lemma 1).¹⁰ In addition, we show that on sparsely connected domains without restoration, any SCF that satisfies swap monotonicity has a special invariance property (Lemma 2). Then, we use Lemmas 1 and 2 to prove that no single agent can influence the outcome by reporting a non-top-adjacent preference (Lemma 4). Finally, we use the results above to show that if two preference profiles with common top alternatives induce different outcomes of f, then f is not unanimous.

Definition 15

(Mishra 2016) An SCF f satisfies swap monotonicity if for any $i \in N$ and $P, P' \in \mathcal {D}$ such that $A(P,P')=\{a,b\}\subset A$, we have for any $P_{-i} \in \mathcal {D}^{n-1}$ that $f(P',P_{-i})=f(P,P_{-i})$ if $f(P,P_{-i}) \notin \{a,b\}$ and $f(P',P_{-i}) \in \{a,b\}$ if $f(P,P_{-i}) \in \{a,b\}$.¹¹

Lemma 1

(Mishra 2016) Let $f:\mathcal {D}^n\rightarrow A$ be a G-LOBIC SCF where $\mathcal {D}\subset \mathcal {P}$. Then, f satisfies swap monotonicity.

Lemma 2

Let $f:\mathcal {D}^n\rightarrow A$ be an SCF where $\mathcal {D}$ is a sparsely connected domain without restoration. If f satisfies swap monotonicity, then it has the following property. For any agent $i \in N$, adjacent preferences $P,P' \in \mathcal {D}$, and a preference profile $P_{-i} \in \mathcal {D}^{n-1}$ such that $A(P,P')=\{a,b\}$ for some $a,b\in A$ where $\{a,b\}\cap \bar{A}\ne \varnothing$, if $f(P,P_{-i})\ne f(P',P_{-i})$, then $f(P,P_{-i}')=f(P,P_{-i})$ and $f(P', P_{-i}')=f(P',P_{-i})$ for any $P_{-i}' \in \mathcal {D}^{n-1}$ such that $a \mathrel {P_{j}'} b$ if and only if $a \mathrel {P_j} b$ for every $j \in N\setminus \{i\}$.

Proof

Without loss of generality (w.l.o.g.), assume that $f(\hat{P},\hat{P}_{-1})\ne f(\hat{P}',\hat{P}_{-1})$ for some adjacent preferences $\hat{P},\hat{P}' \in \mathcal {D}$ such that $A(\hat{P},\hat{P}') = \{a,b\} \subset A$ and some profile $\hat{P}_{-1}\in \mathcal {D}^{n-1}$. Then, by swap monotonicity, $\{f(\hat{P},\hat{P}_{-1}),f(\hat{P}',\hat{P}_{-1})\}=\{a,b\}$. Assume w.l.o.g. that $f(\hat{P},\hat{P}_{-1})=a$ and $f(\hat{P}',\hat{P}_{-1})=b$. Let $\tilde{P}$ denote the preference of agent 2 in $\hat{P}_{-1}$ and assume w.l.o.g. that $a \mathrel {\tilde{P}} b$. Let $\tilde{P}'\in \mathcal {D}$ be another preference such that $a \mathrel {\tilde{P}'}b$. Construct a path in $\mathcal {D}$ without $\{a,b\}$-restoration from $\tilde{P}$ to $\tilde{P}'$:¹²

$$\begin{aligned} (\tilde{P}=P^1,P^2,\ldots ,P^{h-1},P^{h}=\tilde{P}').^{12} \end{aligned}$$

For simplicity, for any $l\in \{1,\ldots ,h\}$, let $\tilde{f}(P^l)\equiv f(\hat{P}, P^l,\hat{P}_{-\{1,2\}})$ and $\tilde{f'}(P^l)\equiv f(\hat{P}', P^l,\hat{P}_{-\{1,2\}})$. We argue that for any $l\in \{1,\ldots ,h-1\}$,

$$\begin{aligned}[\tilde{f}(P^{l})=a \text { and } \tilde{f'}(P^{l})=b] \Rightarrow [\tilde{f}(P^{l+1})=a \text { and } \tilde{f'}(P^{l+1})=b].\end{aligned}$$

Fix $l \in \{1,\ldots ,h-1\}$. Since the path is without $\{a,b\}$-restoration, $A(P^l,P^{l+1})\ne \{a,b\}$. Consider two possible cases: $A(P^l,P^{l+1})\cap \{a,b\}=\varnothing$ and $A(P^l,P^{l+1})\cap \{a,b\}\ne \varnothing$. If it is the former case, then by swap monotonicity, $\tilde{f}(P^{l+1})=a$ and $\tilde{f'}(P^{l+1})=b$. In the latter case, assume w.l.o.g. that $A(P^l,P^{l+1})=\{a,c\}$ for some $c\in A$ such that $c\ne b$. Then, $\tilde{f}(P^{l+1}) \in \{a,c\}$ and $\tilde{f'}(P^{l+1})=b$ by swap monotonicity. However, if $\tilde{f}(P^{l+1})=c$, then $\tilde{f'}(P^{l+1})=c$ since $A(\hat{P},\hat{P}')=\{a,b\}$, which is a contradiction. Therefore, $\tilde{f}(P^{l+1})=a$ and $\tilde{f'}(P^{l+1})=b$. Since $\tilde{f}(\tilde{P})=a$ and $\tilde{f'}(\tilde{P})=b$, $\tilde{f}(\tilde{P}')=a$ and $\tilde{f'}(\tilde{P}')=b$. This shows that $f( \hat{P},P,\hat{P}_{-\{1,2\}}) = a$ and $f(\hat{P}',P,\hat{P}_{-\{1,2\}}) = b$ for any $P\in \mathcal {D}$ such that $a \mathrel {P} b$.

Finally, starting from any pair of profiles $(\hat{P},P,\hat{P}_{-\{1,2\}})$ and $(\hat{P}',P,\hat{P}_{-\{1,2\}})$ such that $a \mathrel {P} b$, we can apply the same reasoning to agents from 3 to n to complete the proof. $\square$

Lemma 3

Let $f:\mathcal {D}^n \rightarrow A$ be an SCF where $\mathcal {D}$ is a sparsely connected domain without restoration. If f is unanimous and G-LOBIC, then $f({\varvec{P}}) \in \bar{A}$ for any ${\varvec{P}} \in \mathcal {D}^n$.

Proof

Suppose not, that is, there exists ${\varvec{P}} \in \mathcal {D}^n$ such that $f({\varvec{P}}) \notin \bar{A}$. Consider a path from one agent’s preference in ${\varvec{P}}$ to one fixed preference. Continue the same process until we find the following agent and adjacent preferences. By unanimity and swap monotonicity, there exist an agent in N, say, agent 1, adjacent preferences $\bar{P},\bar{P}' \in \mathcal {D}$, and a profile $\bar{P}_{-1} \in \mathcal {D}^{n-1}$ such that $\bar{P}(1)=\bar{P}'(1)=a^*$, $A(\bar{P},\bar{P}') = \{a,b\}$, $f(\bar{P},\bar{P}_{-1}) = a$, and $f(\bar{P}',\bar{P}_{-1}) = b$ for some $a^*,a\in \bar{A}$ and $b\notin \bar{A}$, where $a^*$ can be a.

Now, we construct a preference profile $\tilde{P}_{-1}$ with $\tilde{P}_j = a^*,\forall j\in \{2\ldots ,n\}$ such that $f(\bar{P},\tilde{P}_{-1}) \ne a^*$, which contradicts to the unanimity of f.

First, modify the preference of agent 2, fixing the preferences of agents 3 and above, in the following way. (For brevity, let $\bar{f}(P)\equiv f(\bar{P},P,\bar{P}_{-\{1,2\}})$ and $\bar{f'}(P)\equiv f(\bar{P}',P,\bar{P}_{-\{1,2\}})$ for each $P \in \mathcal {D}$.)

Case 1. If $\bar{P}_2(1)=a^*$, take $\tilde{P}_2=\bar{P}_2$. Then $\bar{f}(\tilde{P}_2)=a$ and $\bar{f'}(\tilde{P}_2)=b$.

Case 2. If $\bar{P}_2(1) \ne a^*$ and $a \mathrel {\bar{P}_2} b$, take $\tilde{P}_2=\bar{P}$ so that the relative ranking between a and b for $\tilde{P}_2$ matches that of $\bar{P}_2$. Then, $\bar{f}(\tilde{P}_2)=a$ and $\bar{f'}(\tilde{P}_2)=b$ by Lemma 2.

Case 3. If $\bar{P}_2(1) \ne a^*$ and $b \mathrel {\bar{P}_2} a$, take $\tilde{P}_2 = \bar{P}'$. Then, $\bar{f}(\tilde{P}_2)=a$ and $\bar{f'}(\tilde{P}_2)=b$ by Lemma 2.

Starting from the preference profiles $(\bar{P},\tilde{P}_2,\bar{P}_{-\{1,2\}})$ and $(\bar{P}',\tilde{P}_2,\bar{P}_{-\{1,2\}})$, we can apply the same procedure to the preference orderings of agents from 3 to n. Letting $\tilde{P}_{-1} = (\tilde{P}_2,\ldots ,\tilde{P}_n)$, this yields $f(\bar{P},\tilde{P}_{-1})=a \ne a^*$ and $f(\bar{P}',\tilde{P}_{-1}) = b \ne a^*$, where $\tilde{P}_{j}(1)=a^*$ for all $j \in \{2,\ldots ,n\}$. Since $\bar{P}(1)=\bar{P}'(1)=a^*$, this contradicts unanimity of f. $\square$

Lemma 4

Let $f: \mathcal {D}^n \rightarrow A$ be an SCF where $\mathcal {D}$ is a sparsely connected domain without restoration. If f is unanimous and G-LOBIC, then for any agent $i\in N$, preference $P\in \mathcal {D}$, and preference profile $P_{-i}\in \mathcal {D}^{n-1}$, $f(P',P_{-i})=f(P,P_{-i})$ for any non-top-adjacent preference $P'$ of P.

Proof

Suppose not, i.e., for some agent, say, agent 1, $f(\hat{P}',\hat{P}_{-1})\ne f(\hat{P},\hat{P}_{-1})$ for some adjacent preferences $\hat{P},\hat{P}'$ with $a^*\equiv \hat{P}(1)=\hat{P}'(1)$, $A(\hat{P},\hat{P}')=\{a,b\}$, $a^* \notin \{a,b\}$, and $a \mathrel {\hat{P}} b$, and some preference profile $\hat{P}_{-1}\in \mathcal {D}^{n-1}$. Then, by swap monotonicity, either [$f(\hat{P},\hat{P}_{-1})=a$ and $f(\hat{P}',\hat{P}_{-1})=b$] or [$f(\hat{P},\hat{P}_{-1})= b$ and $f(\hat{P}',\hat{P}_{-1})=a$] holds. W.l.o.g., assume the former case.

As in the proof of Lemma 3, modify the preference orderings of agents from 2 to n, to either $\hat{P},\hat{P}'$. In specific, for any $j \ne i$, let $\tilde{P}_j = \hat{P}$ if $a \mathrel {\hat{P}_j} b$ and $\tilde{P}_j=\hat{P}'$ if $b \mathrel {\hat{P}_j} a$. Then, $f(\hat{P},\tilde{P}_{-1}) = a$ and $f(\hat{P}',\tilde{P}_{-1}) = b$ by Lemma 2, which contradicts unanimity of f. (Note that Lemma 2 can be applied since $f(\hat{P},\hat{P}_{-1}) =a$, $a \in \bar{A}$ by Lemma 3.) $\square$

Now, we use the Lemmas above to show that for any agent $i \in N$, preferences $P,P' \in \mathcal {D}$ with $P(1)=P'(1)$, and preference profile $P_{-i} \in \mathcal {D}^n$, $f(P,P_{-i})=f(P',P_{-i})$ holds.

Fix an agent $i \in N$, preferences $\bar{P},\bar{P}' \in \mathcal {D}$ with $\bar{P}(1)=\bar{P}'(1)=a^*$, and a preference profile $\bar{P}_{-i} \in \mathcal {D}^n$. First, assume that $f(\bar{P},\bar{P}_{-i})\ne a^*$. Since $\mathcal {D}$ is sparsely connected without restoration, there exists a path in $\mathcal {D}$ without $\{a^*,f(\bar{P},\bar{P}_{-i})\}$-restoration connecting $\bar{P}$ and $\bar{P}'$:

$$\begin{aligned}(\bar{P}=P^1,P^2,\ldots ,P^{h-1},P^{h}=\bar{P}').\end{aligned}$$

If $P^l(1)=a^*$ for every $l \in \{1,\ldots ,h\}$, that is, alternative $a^*$ is never swapped in the sequence, we can apply Lemma 4 to every swap of the sequence so that $f(\bar{P}',\bar{P}_{-i})=f(\bar{P},\bar{P}_{-i})$. On the other hand, suppose that there exists $l'\in \{1,\ldots ,h\}$ such that $P^{l'}(1)\ne a^*$.¹³ By swap monotonicity and Lemma 4, for any $l\in \{1,\ldots , h-1\}$, $f(P^l,\bar{P}_{-i})\ne f(P^{l+1},\bar{P}_{-i})$ is only possible when $A(P^l,P^{l+1})=\{P^l(1),P^{l+1}(1)\}$ and $f(P^l,\bar{P}_{-i})\in A(P^l,P^{l+1})$. However, since the path is without $\{a^*,f(\bar{P},\bar{P}_{-i})\}$-restoration, we have

$a^* \mathrel {P^l} f(\bar{P},\bar{P}_{-i})$ for every $l\in \{1,\ldots ,h\}$, which implies that $f(\bar{P},\bar{P}_{-i})$ cannot be the top for every preference in the path. Therefore, $f(P^{l+1},\bar{P}_{-i})= f(P^l, \bar{P}_{-i})$ for every $l \in \{1,\ldots ,h-1\}$ and thus, $f(\bar{P}',\bar{P}_{-i})=f(\bar{P},\bar{P}_{-i})$.

Finally, assume that $f(\bar{P},\bar{P}_{-i})=a^*$. A similar argument from above shows that$f(\bar{P}',\bar{P}_{-i}) = a^*$. $\square$

Proof of Proposition 2

We show that for any profile of generic priors $\{\mu _i\}_{i\in N}$, f is LOBIC with respect to $\{\mu _i\}_{i\in N}$ if and only if it is OBIC with respect to $\{\mu _i\}_{i\in N}$.

OBIC implies LOBIC by definition. For the other direction, suppose that f is LOBIC with respect to $\{\mu _i\}_{i\in N}$. Fix an agent $i \in N$, preferences $\bar{P},\bar{P}' \in \mathcal {D}$, and an alternative $a \in A$. Since $\mathcal {D}$ is connected without restoration, there exists a sequence connecting $\bar{P}$ and $\bar{P}'$ without restoration:

$$\begin{aligned}S=(\bar{P}=P^1,P^2,\ldots ,P^{h-1},P^{h}=\bar{P}').\end{aligned}$$

We show that for any $l\in \{2,\ldots ,h-1\}$,

$$\begin{aligned}\pi _i^f(B(a,\bar{P}),P^l) \ge \pi _i^f(B(a,\bar{P}),P^{l+1}).\end{aligned}$$

Fix $l \in \{2,\ldots ,h-1\}$. There are two possible cases, depending on the intersection of $A(P^l, P^{l+1})$ and $B(a,\bar{P})$:

Case 1. Suppose $A(P^l,P^{l+1}) \cap B(a,\bar{P})=\varnothing$ or $A(P^l,P^{l+1}) \subset B(a, \bar{P})$. Then, $\pi _{i}^f (B(a,\bar{P}),P^l) =\pi _{i}^f(B(a,\bar{P}),P^{l+1})$ by swap monotonicity.

Case 2. Suppose $A(P^l,P^{l+1})=\{x,y\}$ for some $\{x,y\} \subset A$ and $\{x,y\}\cap B(a,\bar{P})=\{x\}$, so that $x \mathrel {\bar{P}} y$. Since S is without $\{x,y\}$-restoration, $x \mathrel {P^{l}} y$ holds. Furthermore, since f is LOBIC, $\pi _{i}^f (B(x,P^l),P^l) \ge \pi _{i}^f (B(x,P^l),P^{l+1})$ holds, which implies that

$\pi _i^f (x,P^l) \ge \pi _i^f (x,P^{l+1})$. Moreover, $\pi _i^f (b,P^l)=\pi _i^f (b,P^{l+1})$ holds for any $b \in B(a,\bar{P}) \setminus \{x\}$ by swap monotonicity. Therefore, $\pi _i^f (B(a,\bar{P}),P^l) \ge \pi _i^f (B(a,\bar{P}),P^{l+1})$.

Since $\pi _i^f(B(a,\bar{P}),\bar{P})\ge \pi _i^f(B(a,\bar{P}),P^2)$ holds by LOBIC, we have $\pi _i^f (B(a,\bar{P}),\bar{P}) \ge \pi _i^f (B(a,\bar{P}),\bar{P}')$. $\square$

Proof of Proposition 3

For the equivalence between G-LOBIC and G-OBIC, it suffices to show that G-LOBIC implies G-OBIC. Suppose that f is LOBIC with respect to a profile of generic priors $\{\mu _i\}_{i\in N}$. For a pair of preferences $\bar{P},\bar{P}'\in \mathcal {D}$, consider a path connecting $\bar{P}$ and $\bar{P}'$:

$$\begin{aligned}(\bar{P}=P^1,P^2,\ldots ,P^{h-1},P^{h}=\bar{P}').\end{aligned}$$

For any agent $i\in N$ and any alternative $a\in A$, $\pi _i^f(B(a,\bar{P}),\bar{P})\ge \pi _i^f(B(a,\bar{P}),P^2)$ holds by LOBIC. Proving that the following holds for any integer $l\in \{2,\ldots ,h-1\}$ completes the proof:

$$\begin{aligned} \pi _i^f(B(a,\bar{P}),\bar{P})\ge \pi _i^f(B(a,\bar{P}),P^l) \Rightarrow \pi _i^f(B(a,\bar{P}),\bar{P})\ge \pi _i^f(B(a,\bar{P}),P^{l+1}). \end{aligned}$$

Suppose $\pi _i^f(B(a,\bar{P}),\bar{P})\ge \pi _i^f(B(a,\bar{P}),P^l)$ holds for some $l\in \{1,\ldots ,h-1\}$. If $P^{l+1}(1)=P^l(1)$, then $\pi _i^f(B(a,\bar{P}),P^{l+1})= \pi _i^f(B(a,\bar{P}),P^l)$ since f is tops-only by Proposition 1. If $P^{l+1}(1)\ne P^l(1)$, we need only consider the case where $P^l(1)\notin B(a,\bar{P})$ and $P^{l+1}(1)\in B(a,\bar{P})$ since otherwise, $\pi _i^f(B(a,\bar{P}),P^{l+1}) \le \pi _i^f(B(a,\bar{P}),P^l)$ by swap monotonicity of f (as in the proof of Proposition 2).

Let $x = P^{l+1}(1)$. Since $\mathcal {D}$ is sparsely connected without restoration, for any $y\notin B(a,\bar{P})$, there exists a path from $\bar{P}$ to $P^{l+1}$ without $\{x,y\}$-restoration:

$$\begin{aligned}(\bar{P}=\tilde{P}^1,\tilde{P}^2,\ldots ,\tilde{P}^{\tilde{h}-1},\tilde{P}^{\tilde{h}}=P^{l+1}).\end{aligned}$$

Furthermore, since $x \mathrel {\bar{P}} y$ and $x \mathrel {P^{l+1}} y$, there does not exist $m \in \{2,\ldots ,\tilde{h}-1\}$ such that $\tilde{P}^{m}(1)=y$. Then, tops-onlyness and swap monotonicity and LOBIC of f imply that $\pi _i^f(y,\bar{P})=\pi _i^f(y,\tilde{P}^{2})= \cdots =\pi _i^f(y,\tilde{P}^{\tilde{h}-1}) =\pi _i^f(y,P^{l+1})$. Therefore, $\pi _i^f(A\setminus B(a,\bar{P}),P^{l+1})=\pi _i^f(A\setminus B(a,\bar{P}),\bar{P})$ and thus $\pi _i^f(B(a,\bar{P}),P^{l+1})=\pi _i^f(B(a,\bar{P}),\bar{P})$.

For the equivalence between LDSIC and DSIC, it suffices to show that LDSIC implies DSIC. Suppose, on the contrary, that there exists an agent, say, agent 1, preferences $\bar{P},\bar{P}'\in \mathcal {D}$, and a preference profile $\bar{P}_{-1} \in \mathcal {D}^{n-1}$ such that $f(\bar{P}',\bar{P}_{-1}) \mathrel {\bar{P}} f(\bar{P},\bar{P}_{-1})$. For simplicity, let $x = f(\bar{P}',\bar{P}_{-1})$ and $y = f(\bar{P},\bar{P}_{-1})$. If $x \mathrel {\bar{P}'} y$, every path connecting $\bar{P}$ and $\bar{P}'$ is with $\{x,y\}$-restoration since tops-onlyness and LDSIC of f imply that for some preference $\hat{P}$ in the path, $\hat{P}(1)=y$. This contradicts the assumption that $\mathcal {D}$ is sparsely connected without restoration. Therefore, we can assume that $y \mathrel {\bar{P}'} x$. Consider a path $(\bar{P}=P^1,P^2,\ldots ,P^{h-1},P^{h}=\bar{P}')$ from $\bar{P}$ to $\bar{P}'$ without $\{x,y\}$-restoration. Then, there exists $l\in \{1,\ldots ,h-1\}$ such that $A(P^l,P^{l+1})=\{a,y\}$ and $P^l(1)=a$ for some $a \in A$, $f(P^l,\bar{P}_{-1})=y$, and $f(P^{l+1},\bar{P}_{-1})=a$ (a could be x). Since $a \mathrel {P^l} y$, this contradicts that f is LDSIC. $\square$

Proof of Proposition 4

Connectedness Assume that the domain is not connected. Then, there exists a pair of preferences $\bar{P},\bar{P}' \in \mathcal {D}$ for which there is no connecting path in $\mathcal {D}$. Let $\mathcal {D}^{\bar{P}} \subset \mathcal {D}$ be the set of preferences that is connected to $\bar{P}$ in $\mathcal {D}$, including $\bar{P}$. Since $\bar{P}' \notin \mathcal {D}^{\bar{P}}$, $\mathcal {D}\setminus \mathcal {D}^{\bar{P}}$ is nonempty.

We construct an SCF f that is unanimous and LDSIC but not dictatorial:

$$\begin{aligned} f(\varvec{P})={\left\{ \begin{array}{ll} P_1(1) &{} \text {if } P_{1} \in \mathcal {D}^{\bar{P}}\\ P_2(1) &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

f is unanimous since it always chooses the top alternative of either agent 1 or 2. f is not dictatorial since $|\bar{A}| \ge 2$. To see that f is LDSIC, we need only check the incentive constraints of agent 1. For any $P \notin \mathcal {D}^{\bar{P}}$, agent 1 can only change the outcome of f if he reports a preference in $\mathcal {D}^{\bar{P}}$. However, P is not adjacent to any preference in $\mathcal {D}^{\bar{P}}$, since otherwise, there exists a path from $\bar{P}$ to P, which contradicts that $P \notin \mathcal {D}^{\bar{P}}$.

Disagreement property Assume that the domain does not satisfy the disagreement property. Then, there exist top-adjacent preferences $\bar{P},\bar{P}'$ such that $\bar{P}(1)=a$, $\bar{P}'(1)=b$, and either [$a \mathrel {P} b,\forall P\in \mathcal {D}\setminus \mathcal {D}^b$] or [$b \mathrel {P} a,\forall P\in \mathcal {D}\setminus \mathcal {D}^a$] holds. W.l.o.g., assume that $b \mathrel {P} a,\forall P\in \mathcal {D}\setminus \mathcal {D}^a$. Then, for any adjacent preferences $P,P'$ such that $P(1)=a$, $P'(1) \in \{a,b\}$ holds since otherwise, $P'(2)=a$, which implies $a \mathrel {P'} b$.

We construct an SCF f that is unanimous and LDSIC but not dictatorial:

$$\begin{aligned} f(\varvec{P})={\left\{ \begin{array}{ll} b &{} \text {if } P_{1}(1)=a \text { and } P_{2}(1)\ne a\\ P_{1}(1) &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

It is straightforward to see that f is unanimous and non-dictatorial. To show that f is LDSIC, it is sufficient to check the incentive constraints of agent 1 and 2. For agent 1, when $P_{1}(1)=a$ and $P_{2}(1)\ne a$, adjacent preferences that agent 1 can report either have a or b as the top alternative. For agent 2, consider three cases when $P_1(1)=a$:

Case 1. If $P_{2}(1)=a$, $f({\varvec{P}})=a$, so there is no incentive to lie.

Case 2. If $P_{2}(1)=b$, $f({\varvec{P}})=b$. Again, there is no incentive to lie.

Case 3. If $P_{2}(1)\notin \{a,b\}$, $f({\varvec{P}})=b$. Since $bP_2a$, agent 2 does not have an incentive to lie. $\square$

Proof of Theorem 1, 1 $\Rightarrow$ 2

Lemma 5

If $\mathcal {D}$ is sparsely connected without restoration and satisfies the disagreement property, there exist a pair of top-adjacent preferences $\bar{P},\bar{P}' \in \mathcal {D}$ such that at least one pair of their associated preferences has the same top alternative other than $\bar{P}(1)$ or $\bar{P}'(1)$.

Proof

Assume to the contrary that for every pair of top-adjacent preferences, all pairs of associated preferences have different top alternatives. Suppose that $\bar{P}$ and $\bar{P}'$ are top-adjacent preferences with $\bar{P}(1)=a$ and $A(\bar{P},\bar{P}') = \{a,b\}$. Then, we can partition the set of top alternatives $\bar{A}$ into two sets $\bar{A}^a$ and $\bar{A}^b$, where $\bar{A}^a = \{c\in \bar{A} : a \mathrel {P} b, \forall P \text { s.t. } P(1) = c\}$ and $\bar{A}^b = \{c\in \bar{A} : b \mathrel {P} a, \forall P \text { s.t. } P(1) = c\}$. Note that both are nonempty since $a \in \bar{A}^a$ and $b \in \bar{A}^b$. The domain $\mathcal {D}$ can also be partitioned in a similar way: $\mathcal {D} = \mathcal {D}^a \cup \mathcal {D}^b$ where $\mathcal {D}^a = \{P \in \mathcal {D}: P(1) \in \bar{A}^a \}$ and $\mathcal {D}^b = \{P \in \mathcal {D}: P(1) \in \bar{A}^b \}$.

Consider a path starting from $\bar{P}(=P^1)$ constructed as follows:

Step 1. Select any preference ordering $\hat{P} \in \mathcal {D}^a$ such that $\hat{P}(1) \ne a$ and let $P^2=\hat{P}$. Construct a path connecting $P^1$ and $P^2$ without $\{a,b\}$-restoration:

$$\begin{aligned} (P^1=\tilde{P}^{1},\tilde{P}^{2},\tilde{P}^{3},\ldots ,\tilde{P}^{k_1-1},\tilde{P}^{k_1}=P^2) \text {.} \end{aligned}$$

Note that all preferences in the path are in $\mathcal {D}^a$ since the path is without $\{a,b\}$-restoration. If there exists k such that $1< k < k_1$ and $\tilde{P}^{k}(1) = P^2(1)$, let $P^2 = \tilde{P}^k$ and take the subpath from $\tilde{P}^1$ to $\tilde{P}^k$, letting $k_1=k$. If $\bar{A}^a = \cup _{j=1}^{k_1} \tilde{P}^{j}(1)$, stop. If not, continue to the next step.

Step 2. Select any preference ordering $\hat{P}' \in \mathcal {D}^a$ such that $\hat{P}'(1) \notin \cup _{j=1}^{k_1} \{\tilde{P}^{j}(1)\}$ and let $P^3 = \hat{P}'$. Construct a path connecting $P^2$ and $P^3$ without $\{a,b\}$-restoration.

$$\begin{aligned} (P^2=\tilde{P}^{k_1+1},\tilde{P}^{k_1+2},\ldots ,\tilde{P}^{k_2-1},\tilde{P}^{k_2}=P^3) \text {.} \end{aligned}$$

If there exists k such that $k_1< k < k_2$ and $\tilde{P}^{k}(1) = P^3(1)$, let $P^3 = \tilde{P}^k$ and take the subpath from $\tilde{P}^2$ to $\tilde{P}^k$, letting $k_2=k$. If $\bar{A}^a = \cup _{j=1}^{k_2} \tilde{P}^{j}(1)$, stop. If not, proceed to the next step.

Repeat this process until the union of all top alternatives in the path is equal to $\bar{A}^a$. Since there are a finite number of alternatives, it ends in finite steps. In addition, a similar path can be constructed starting from $\bar{P}'$, using $\bar{A}^b$ and $\mathcal {D}^b$ in place of $\bar{A}^a$ and $\mathcal {D}^a$. Then, connect the two paths by $\bar{P}$ and $\bar{P}'$ as shown in Table 6.

Table 6 Path construction in Lemma 5

Suppose the path construction starting from $\bar{P}$ ends in M steps. Suppose $\tilde{P}^{k_M-1}(1)=x$ and $\tilde{P}^{k_M}(1)=y$. It is easy to see that $P^M$ is the only preference in the sequence whose top alternative is y. In addition, since x and y are top-swapped, by assumption, there does not exist adjacent preferences $\hat{P}$ and $\hat{P}'$ such that $\hat{P}(1)=\hat{P}'(1)$ and $A(\hat{P}, \hat{P}') = \{x,y\}$. This implies that x is strictly preferred to y for all the preference orderings in the path, except for $P^M$. However, since $\mathcal {D}$ satisfies the disagreement property, there exists a preference $P \in \mathcal {D}$ such that $P(1)\ne y$ and $y \mathrel {P} x$. Since the path includes preferences with every top alternative, there exists another preference $P'$ in the path such that $P'(1)=P(1)$ and $x \mathrel {P'} y$, which is a contradiction to the assumption. $\square$

Considering ${D}^a$ and ${D}^b$, we show by the proof of Lemma 5 that if $\bar{P}$ and $\bar{P}'$ do not have any pair of associated preferences with a common top alternative, then there exist two pairs of preferences, $(\hat{P},\hat{P}')$ and $(\tilde{P},\tilde{P}')$ such that

• ($\hat{P},\hat{P}'$) and $(\tilde{P},\tilde{P}')$ are top-adjacent preferences, respectively, such that $(A(\hat{P},\hat{P}') \cap \{a,b\}) = (A(\tilde{P},\tilde{P}') \cap \{a,b\}) = \varnothing$,

• $a \mathrel {\hat{P}} b, a \mathrel {\hat{P}'} b, b \mathrel {\tilde{P}} a, b \mathrel {\tilde{P}'} a$, and

• there exist associated preferences of $(\hat{P},\hat{P}')$ and $(\tilde{P},\tilde{P}')$, respectively, with same tops.

Now, we are ready to prove Theorem 1, 1 $\Rightarrow$ 2.

Step I. By Lemma 5, we have top-adjacent preferences $\bar{P},\bar{P}'$ such that $\bar{P}(1)=a,$ $\bar{P}'(1)=b,$ with associated preferences $\hat{P},\hat{P}'$ such that $\hat{P}(1)= \hat{P}'(1)\notin \left\{ a,b\right\}$, $a\mathrel {\hat{P}}b$, and $b\mathrel {\hat{P}'}a$.

By unanimity, $f(\bar{P},\bar{P},\ldots ,\bar{P})=a$ and $f(\bar{P}',\bar{P}',\ldots ,\bar{P}')=b$. From the preference profile $(\bar{P},\bar{P},.\ldots ,\bar{P})$, consider changing agents’ preferences, one by one, from $\bar{P}$ to $\bar{P}'$. Then, there exists an agent i such that $f(\bar{P},\bar{P}_{-i})=a$ and $f(\bar{P}',\bar{P}_{-i})=b$ for some $\bar{P}_{-i} \in \mathcal {D}^{n-1}$ such that $\bar{P}_j \in \{\bar{P},\bar{P}'\}$ for all $j \ne i$. Fix such an agent i, any such a profile $\bar{P}_{-i}$ and another agent $j \ne i$. W.l.o.g., assume $\bar{P}_j = \bar{P}$. By Lemma 2, $f(\bar{P},\hat{P},\bar{P}_{-\{i,j\}})=a$ and $f(\bar{P}', \hat{P},\bar{P}_{-\{i,j\}})=b$ hold. In addition, since f is tops-only by Proposition 1, $f(\bar{P}, \hat{P}',\bar{P}_{-\{i,j\}})=a$ and $f(\bar{P}',\hat{P}',\bar{P}_{-\{i,j\}})=b$ hold. Combining these two observations and using Lemma 2, we can see that for any preference $P \in \mathcal {D}$, $f(\bar{P},P,\bar{P}_{-\{i,j\}})=a$ and $f(\bar{P}',P,\bar{P}_{-\{i,j\}})=b$ hold.

Applying this argument to every agent other than i and j and using tops-onlyness of f shows that for any $P_{-i} \in \mathcal {D}^{n-1}$, $f(P,P_{-i})=a$ and $f(P',P_{-i})=b$ hold for any $P,P' \in \mathcal {D}$ such that $P(1)=a$ and $P'(1)=b$.

Step II. For any other top-adjacent preferences $\tilde{P},\tilde{P}'$ such that $\tilde{P}(1)=c$ and $\tilde{P}'(1)=d$, first consider the case where there exist associated preferences with same tops. By Step I, there exists an agent $i' \in N$ such that $f(\tilde{P},P_{-i'}) = c$ and $f(\tilde{P}',P_{-i'})=d$ for any profile $P_{-i'} \in \mathcal {D}^{n-1}$. Then, $i'=i$ since if not, $f(\tilde{P},\bar{P},P_{-\{i',i\}}) = c$ for any $P_{-\{i',i\}} \in \mathcal {D}^{n-1}$, which is a contradiction to Step I.

Second, assume that there does not exist a pair of associated preferences to $(\tilde{P},\tilde{P}')$ with same tops. As in previous steps, $f(\tilde{P},\ldots ,\tilde{P}) = c$, $f(\tilde{P}',\ldots ,\tilde{P}') = d$, and there exist an agent $i' \in N$ and a profile $\tilde{P}_{-i'} \in \mathcal {D}^{n-1}$ such that $f(\tilde{P},\tilde{P}_{-i'}) = c$, $f(\tilde{P}',\tilde{P}_{-i'})=d$, and $\tilde{P}_j \in \{\tilde{P},\tilde{P}'\}$ for any $j \ne i'$. First, we argue that $i'=i$. Suppose not, and assume w.l.o.g. that $a \mathrel {\tilde{P}_i} b$. Then, by Lemma 2, $f(\tilde{P},\bar{P},\tilde{P}_{-\{i',i\}}) = c$ holds, which is a contradiction to Step I.

Now, assume w.l.o.g. that $c \mathrel {\bar{P}} d$ and fix an agent $j \ne i$. If $f(\tilde{P},\tilde{P},\tilde{P}_{-\{i,j\}}) \ne f(\tilde{P},\tilde{P}',\tilde{P}_{-\{i,j\}})$, then $f(\bar{P},\tilde{P},\tilde{P}_{-\{i,j\}}) \ne f(\bar{P},\tilde{P}',\tilde{P}_{-\{i,j\}})$ by Lemma 2, which is a contradiction to Step I. This implies that $f(\tilde{P},\tilde{P},\tilde{P}_{-\{i,j\}}) = f(\tilde{P},\tilde{P}',\tilde{P}_{-\{i,j\}}) = c$. To show that $f(\tilde{P}',\tilde{P},\tilde{P}_{-\{i,j\}}) = f(\tilde{P}',\tilde{P}',\tilde{P}_{-\{i,j\}}) = d$, suppose to the contrary, w.l.o.g., that $f(\tilde{P}',\tilde{P},\tilde{P}_{-\{i,j\}}) = c$ and $f(\tilde{P}',\tilde{P}',\tilde{P}_{-\{i,j\}}) = d$. Then, $f(P,\tilde{P},\tilde{P}_{-\{i,j\}}) = c$ and $f(P,\tilde{P}',\tilde{P}_{-\{i,j\}}) =d$ hold for any $P \in \mathcal {D}$ such that $d \mathrel {P} c$. However, by Lemma 5, there exist top-adjacent preferences $\dot{P},\dot{P}' \in \mathcal {D}$ such that $d \mathrel {\dot{P}} c$, $\dot{P}(1)=e$, and $\dot{P}'(1)=f$ where $e,f\notin \{c,d\}$, to whom there exist associated preferences with the same top alternative. By Step I, $f(\dot{P},\tilde{P},\tilde{P}_{-\{i,j\}}) = e$, a contradiction.

By Lemma 2, $f(\tilde{P},P,\tilde{P}_{-\{i,j\}}) = c$ and $f(\tilde{P}',P,\tilde{P}_{-\{i,j\}}) = d$ for any $P \in \mathcal {D}$. Since the same argument can be applied to agents other than i or j, we have $f(\tilde{P},P_{-i}) = c$ and $f(\tilde{P}',P_{-i}) = d$ for any $P_{-i} \in \mathcal {D}^{n-1}$. $\square$

Proof of Theorem 1, 2 $\Rightarrow$ 3 This is immediate since DSIC implies G-LOBIC by definition. $\square$

Proof of Theorem 1, 3 $\Rightarrow$ 1 By Proposition 4, if any unanimous and LDSIC SCF is dictatorial, then $\mathcal {D}$ satisfies the disagreement property. By Proposition 3, this extends to unanimous and DSIC SCFs in SCD. $\square$

Proof of Proposition 5

The proof proceeds as follows: Lemma 6 (a result from Mishra 2016) shows that for any preference domain, G-LOBIC combined with elementary monotonicity is equivalent to LDSIC, which is in turn equivalent to DSIC by Proposition 3. Finally, proving that a unanimous G-LOBIC SCF satisfies weak elementary monotonicity if and only if it satisfies elementary monotonicity completes the proof (Lemma 7).

Lemma 6

(Mishra 2016) For any preference domain $\mathcal {D}\subset \mathcal {P}$, an SCF $f:\mathcal {D}\rightarrow A$ is LDSIC if and only if it is G-LOBIC and satisfies elementary monotonicity.

Lemma 7

Let $f:\mathcal {D}^n\rightarrow A$ be a unanimous and G-LOBIC SCF where $\mathcal {D}$ is a sparsely connected domain without restoration. Then, f satisfies elementary monotonicity if and only if it satisfies weak elementary monotonicity.

Proof of Lemma 7

Elementary monotonicity implies weak elementary monotonicity by definition. To show the other implication, suppose to the contrary that there exist an agent, say, agent 1, adjacent preferences $\bar{P},\bar{P}' \in \mathcal {D}$, and a profile $\bar{P}_{-1} \in \mathcal {D}^{n-1}$ such that $f(\bar{P},\bar{P}_{-1}) \ne a$ and $f(\bar{P}',\bar{P}_{-1})=a$, where $A(\bar{P},\bar{P}') = \{a,b\}$.

Since f is tops-only by Proposition 1, $\bar{P}(1)=a$ and $\bar{P}'(1)=b$. In addition, $f(\bar{P},\bar{P}_{-1})=b$ by swap monotonicity. Let $\bar{P}_j$ denote the preference ordering of agent j in $\bar{P}_{-1}$. Now, we modify the preference orderings for all other agents, each from $\bar{P}_j$ to $\tilde{P}_j \in \{\bar{P},\bar{P}'\}$.

If $\bar{P}_2 \in \{\bar{P},\bar{P}'\}$, take $\tilde{P}_{2}=\bar{P}_{2}$. Otherwise, take $\tilde{P}_{2}=\bar{P}$ if $a \mathrel {\bar{P}_2} b$ and $\tilde{P}_{2}=\bar{P}'$ if $b \mathrel {\bar{P}_2} a$. In both cases, we have $f(\bar{P},\tilde{P}_2,\bar{P}_{-\{1,2\}}) =b$ and $f(\bar{P}',\tilde{P}_{2},\bar{P}_{-\{1,2\}})=a$ by Lemma 2.

Applying the same procedure to agents 3 through n and letting $\tilde{P}_{-1} = (\tilde{P}_2,\tilde{P}_3,\ldots ,\tilde{P}_n)$ gives two top-two profiles $(\bar{P},\tilde{P}_{-1})$ and $(\bar{P}',\tilde{P}_{-1})$ that do not satisfy elementary monotonicity. By definition, this contradicts the weak elementary monotonicity of f. $\square$

$\square$