Equivalent definition of homeomorphic spaces I have a question about homeomorphic spaces. Two topological spaces $(X, τ_{x})$ and $(Y, τ_{y})$ are called homeomorphic iff there exists a bijection $f:X \rightarrow Y$ such that both $f$ and $f^{-1}$ are continuous with respect to the topologies $τ_{x}$ and $τ_{y}$. Is this statement equivalent to the following? 

There exists a bijection $φ: τ_{x} \rightarrow τ_{y}$ such that for all $u, v, u_{i} \in τ_{x}$:

*

*$ φ(\emptyset) = \emptyset, \space φ(X) = Y $

*$ φ(\bigcup_{i \in I} u_{i}) = \bigcup_{i \in I} φ(u_{i}) $

*$ φ(u \cap v) = φ(u) \cap φ(v) $
I don't know allmost anything about topology so I might be very wrong but this seems to me like a very intuitive and "clean" way to define an isomorphism between spaces (treating the union and intersection as operations preserved by $φ$).
Are the two statements equivalent? If not what's a simple counterexample? If yes why don't we use the second (in my opinion simpler) definition? Thanks in advance!
 A: An even more concrete counter example:
Consider $\mathbb{N}_1=\{1\}$ and $\mathbb{N}_3=\{1,2,3\}$ both with the indiscrete topology, so $$\mathcal{T}_{\mathbb{N}_1}=\{\emptyset,\{1\}\}\text{ and }\mathcal{T}_{\mathbb{N}_3}=\{\emptyset,\{1,2,3\}\}$$
Consider the function $\varphi:\mathcal{T}_{\mathbb{N}_1}\to\mathcal{T}_{\mathbb{N}_3}$ given by $$\varphi=\left\{\big(\emptyset,\emptyset\big),\big(\{1\},\{1,2,3\}\big)\right\}$$
You can verify that $\varphi$ satisfies the conditions you required.
The spaces $(\mathcal{T}_{\mathbb{N}_1},\mathbb{N}_1)$ and $(\mathcal{T}_{\mathbb{N}_3},\mathbb{N}_3)$ cannot be homeomorphic because there is no bijection between the two underlying sets.
More generally, any two sets equipped with the indiscrete topology with different cardinalites will suffice as a counter example because two spaces $X$ and $Y$ equipped with the indiscrete topology are homoeomorphic if and only if they have the same cardinality.
A: The two are not equivalent :
If there is a homeomorphism $f:X\to Y$, then $U\in\tau_X \mapsto f(U) \in \tau_Y$ is a bijection which satisfies all your hypotheses.
However, the converse is not true. Take $X = \mathbb R$ with $\tau_X$ the usual topology and $Y = \mathbb R \sqcup \{0'\}$ be $\mathbb R$ with an extra element $0'$ with the topology :
$$\tau_Y = \Big\{U\subset X\sqcup \{0'\}\Big|U\cap X\in\mathbb \tau_X\quad \text{and}\quad 0'\in U\Leftrightarrow 0\in U \Big\}$$
Then there is a bijection $\phi:\tau_X\to\tau_Y$ which satisfies all three conditions, defined by :
$$\forall U\in\tau_X, \phi(U) = \left\{\begin{array}{cl}U\sqcup \{0'\} & \text{if } 0\in U\\
U & \text{else}
\end{array}\right.$$
However, $X$ and $Y$ are not homeomorphic, since $X$ is Hausdorff and $Y$ is not.
