When does a bijection of topologies induce a homeomorphism of spaces? If two topological spaces $(X, \tau)$ and $(Y, \tau')$ are homeomorphic, we have a bijective correspondence between $\tau$ and $\tau'$ via $U \in \tau \mapsto f(U) \in \tau'$ where $f: X \to Y$ is a homeomorphism. 
Are the reasonable conditions to be imposed to a bijection $\Gamma : \tau \to \tau'$ so that it implies that $X$ and $Y$ are homeomorphic? I'm not necessarily asking for a homeomorphism $g : X \to Y$ to verify $\Gamma(U) = g(U)$, but for conditions on $\Gamma$ that imply the existence of some homeomorphism.

So, let me first impress on you how outrageously weak the mere existence of such a bijection $\Gamma$ is.  It means solely that $X$ and $Y$ have the same number of open sets.  When dealing with infinite sets, a cardinality statement like this says extremely little.  For instance, if $X$ is any infinite separable metric space, then there are exactly $2^{\aleph_0}$ open subsets of $X$.  So for any two infinite separable metric spaces $X$ and $Y$, there exists such a bijection $\Gamma$.  This includes, for instance, every infinite subspace of $\mathbb{R}^n$ for any $n$.
One additional much stronger condition you can impose is that $\Gamma$ is an order-isomorphism (with respect to the inclusion order).  That is, $\Gamma$ is a bijection and for any $U,V\in\tau$, $U\subseteq V$ iff $\Gamma(U)\subseteq \Gamma(V)$.  For nice spaces, this implies that there is a homeomorphism $g:X\to Y$ such that $\Gamma(U)=g(U)$ for all $U\in\tau$.  In particular, if $X$ and $Y$ are both $T_1$, you can recover $g$ by considering open subsets of the form $X\setminus\{x\}$.  These subsets can be characterized in terms of the order relation (for instance, they are the elements of $\tau$ that have exactly one other element of $\tau$ which contains them).  For each $x\in X$, $\Gamma(X\setminus\{x\})$ must therefore be of the form $Y\setminus\{y\}$ for some $y\in Y$.  Defining $g(x)$ to be this $y$, it is then not hard to verify that $\Gamma(U)=g(U)$ for all $U$ and $g$ is a homeomorphism.
(Alternatively, instead of assuming $X$ and $Y$ are both $T_1$, it would suffice to assume they are both sober.)
