Complexity of the search version of 2-SAT assuming $\mathsf{L = NL}$
Given a satisfiable 2-CNF $\phi$, you can compute a particular satisfying assignment $e$ by an NL-function (that is, there is an NL-predicate $P(\phi,i)$ that tells you whether $e(x_i)$ is true). One way to do that is described below. I will freely use the fact that NL is closed under $\mathrm{AC}^0$-reductions, hence NL-functions are closed under composition; this is a consequence of NL = coNL.

Let $\phi(x_1,\dots,x_n)$ be a satisfiable 2-CNF. For any literal $a$, let $a^\to$ be the number of literals reachable from $a$ by a directed path in the implication graph of $\phi$, and $\let\ot\leftarrow a^\ot$ the number of literals from which $a$ is reachable. Both are computable in NL.

Observe that $\let\ob\overline \ob a^\to=a^\ot$, and $\ob a^\ot=a^\to$, due to skew-symmetry of the implication graph. Define an assignment $e$ so that


if $a^\ot&gt;a^\to$, then $e(a)=1$;
if $a^\ot&lt;a^\to$, then $e(a)=0$;
if $a^\ot=a^\to$, let $i$ be minimal such that $x_i$ or $\ob x_i$ appears in the strongly connected component of $a$ (it cannot be both, as $\phi$ is satisfiable). Put $e(a)=1$ if $x_i$ appears, and $e(a)=0$ otherwise.


The skew-symmetry of the graph implies that $e(\ob a)=\ob{e(a)}$, hence this is a well-defined assignment. Moreover, for any edge $a\to b$ in the implication graph:


If $a$ is not reachable from $b$, then $a^\ot&lt;b^\ot$, and $a^\to&gt;b^\to$. Thus, $e(a)=1$ implies $e(b)=1$.
Otherwise, $a$ and $b$ are in the same strongly connected component, and $a^\ot=b^\ot$, $a^\to=b^\to$. Thus, $e(a)=e(b)$.


It follows that $e(\phi)=1$.