Homeomorphism via map between topologies

Question

Background

Theorem. Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be two topological spaces and $f:(X,\tau_X)\to(Y,\tau_Y)$ be a homeomorphism between them. Then there exists a bijection $\varphi:\tau_X\to\tau_Y$.

Sketch of the Proof. Define $\varphi:\tau_X\to\tau_Y$ by $\varphi(U)=f(U)$. Then the map is well-defined because $f(U)\in\tau_Y$ for all $U\in\tau_X$. To show that $\varphi$ is injective observe that for all $U,V\in\tau_X$,\begin{align}\varphi(U)=\varphi(V)\implies f(U)=f(V)\tag{1}\end{align}If possible, let $U\not\subseteq V$ or $V\not\subseteq U$. Without loss of generality let us assume that $U\not\subseteq V$. Then there exists $x\in U$ such that $x\notin V$. Then $f(x)\in f(U)$ and $f(x)\notin f(V)$ (since $f$ is injective). Consequently $f(U)\ne f(V)$ contrary to $(1)$. Similarly we can prove that it is impossible to have $V\not\subseteq U$. Consequently $U=V$.

To prove the surjectivity of $\varphi$, let $V\in \tau_Y$. Then observe that since $f$ is continuous $f^{-1}(V)\in\tau_X$ and since $f$ is surjective we have, $$\varphi(f^{-1}(V))=f(f^{-1}(V))=V$$ Consequently $\varphi$ is bijective.

Questions

1. Is the converse of the above theorem true? More specifically, if there exist bijections between $\tau_X,\tau_Y$ and $X,Y$ respectively then are $(X,\tau_X)$ and $(Y,\tau_Y)$ homeomorphic?

2. Let $\phi:\tau_X\to\tau_Y$ be any function. Then does $\phi$ always "induces" an open map from $(X,\tau_X)$ to $(Y,\tau_Y)$?

3. Let $\phi:\tau_Y\to\tau_X$ be any function. Then does $\phi$ always "induces" a continuous function from $(X,\tau_X)$ to $(Y,\tau_Y)$?

My Conjecture

I think that the answer to my first question is most probably affirmative. However, I couldn't construct such a map till now. Regarding the other two I have no idea. Can anyone help (especially regarding the first question by constructing explicitly such map)?

Answer 1

The sets $\tau_X$ and $\tau_Y$ have more structure on them than just bare sets: they are partially ordered sets, where the partial ordering on $\tau_X$ is the inclusion relation on subsets of $X$, and similarly for $\tau_Y$. The bijection $\phi : \tau_X \to \tau_Y$ that is induced by a homeomorphism $f : (X,\tau_X) \to (Y,\tau_Y)$ preserves that partial ordering.

So the answer to all of your questions is no. Assuming $X$ and $Y$ are not empty, and using that $\tau_X$ and $\tau_Y$ have the same cardinality (which is implied by your statement that there exists some bijection between them), it is easy to define a bijection $\phi : \tau_X \to \tau_Y$ that not preserve inclusion. Simply define $\phi(\emptyset)=Y$ and $\phi(X)=\emptyset$, and extend this arbitrarily to a bijection. The same $\phi$ gives a "no" answer to all of your questions.