Are different assignments allowed for the implication graph proof of 2-SAT being in P? One proof for $2-SAT$ being in $P$ uses the implication graph, i.e. one creates 2 vertices per variable $a$, one for each possible literal ($a$ and $\neg a$). One then adds 2 arcs per clause $(a \lor b)$, one from $\neg a$ to $b$ and one from $\neg b$ to $a$, respectively representing the implications $\neg a \implies b$ and $\neg b \implies a$. Finally, (assuming a solution exists) one sorts the connected components in topological order and traverses these in reverse order. Now for the important part :
For every literal $a$ encountered, set variable $a = true$, and for every literal $\neg a$ encountered, set variable $a = false$.
Don't change variables already set. The resulting assignment is a solution.
My question is : Do other assignments work as well, as long as the underlying implications stay $true$ ? For example :
For every literal $a$ encountered, set variable $a = false$, and for every literal $\neg a$ encountered, set variable $a = true$.
since $false \implies false$, or even some combination :
For every literal $a$ encountered, if it points to some literal that equals $false$, set $a = false$, otherwise set it to either $false$ or $true$. For every literal $\neg a$ encountered, if it points to some literal that equals $false$, set $a = true$, otherwise set it to either $false$ or $true$.
since $true \implies true$, $false \implies true$, and $false \implies false$?
You have to understand that $a \implies b$ cannot be interpreted as $\neg a \implies \neg b$. Only the contraposition $b \implies a$ is true, but this one is already in the implication graph by construction.
Thus, with your algorithm, you will make abrirtrary assignments. This will probably leads to an impossibility even if the problem does have a solution. Of course, on trivial examples, one can see that this algorithm may produce a solution.
Also note that  $a \implies b \implies c$ means there is elsewhere in the graph $\neg c \implies \neg b \implies \neg a$. So when you decide to assign false in order to $a, b, c$, you are reading the graph and the strongly connected components reversly.