Let $n$ be the number of voters and $A$ be the set of alternatives. For voter $i$, we denote by $a \succ_i b$, if $i$ prefers $a$ to $b$, where $a,b \in A$. Let $L(A)$ denote the set of all strict linear orders on $A$.
The proof of the Gibbard Satterthwaite theorem starts by considering the case $n=2$.
First part of the proof: Let $a,b \in A$ with $a \neq b$. Consider $\succ \in L(A)^2$ such that \begin{equation} a \succ_1 b \succ_1 x \quad \text{ and } \quad b \succ_2 a \succ_2 x, \end{equation} for all $x \in A \setminus \{a, b \}$. Let $f: (L(A))^2 \rightarrow A$ be a SCF. Suppose, for a contradiction, that $f(\succ) =a.$ We also define $\succ' \in L(A)^2$ such that \begin{equation} a \succ^{'}_1 b \succ^{'}_1 x \quad \text{ and } \quad b \succ^{'}_2 x \succ^{'}_2 a, \end{equation} for all $x \in A \setminus \{a, b \}$. Then it is clear that $f( \succ') \neq b$ by strategy-proofness. ....
My question is that I don't understand why $f( \succ') \neq b$ by strategy-proofness? I can't show that $f$ is manipulable if $f( \succ') = b$. Any ideas?
If $f(\succ')=b$, then $2$ is better off pretending to have $\succ_2'$ rather than the actual $\succ_2$: the outcome will then be $b$ instead of $a$, and $b\succ_2 a$.