Through comparing topologies, one could deduce a more compact notation for continuity. It also leads to a nice description of continuous bijections and homemorphisms.

## Comparison of topologies and continuity

Given two topologies $$\tau_1, \tau_2$$ of $$X$$. We say that $$\tau_1$$ is coarser than $$\tau_2$$ iff $$\tau_1 \subseteq \tau_2$$. We also say that $$\tau_1$$ is finer than $$\tau_2$$ iff $$\tau_2 \subset \tau_1$$. Intuitively a topology is coarser than another if it has more open sets.

This has links with continuity as demonstrated below.

Theorem: $$\text{id}_X: (X, \tau_1) \to (X, \tau_2)$$ is continuous iff $$\tau_2 \subseteq\tau_1$$

As such we could see that $$\text{id}_X: (X, \tau_1) \to (X, \tau_2)$$ is continuous iff $$\tau_2$$ is coarser than $$\tau_1$$.

This idea generalises to arbitrary functions. Note that for every function $$f: X \to Y$$, it induces a function $$f': P(X) \to P(Y)$$, where $$P(X), P(Y)$$ are the power sets of $$X$$ and $$Y$$ such that $$f'(\Omega) := \bigcup_{S \in \Omega}\text{Im}_f(S)$$ for all $$\Omega \in P(X)$$. Depending on context, we shall use $$f'$$ and $$f$$ interchangebaly.

For full generality we see that

Theorem: $$f: (X, \tau_1) \to (X, \tau_2)$$ is continuous iff $$f^{-1}(\tau_2) \subseteq \tau_1$$

Theorem: $$f: (X, \tau_X) \to (Y, \tau_Y)$$ is continuous iff $$f^{-1}(\tau_Y) \subseteq \tau_X$$

This is just a fancy way of saying that $$\; \forall U \in \tau_Y, f^{-1}(U) \in \tau_X$$, the classical definition of continuity.

In fact, given some map $$f: X \to Y$$ where $$Y$$ has some topology $$\tau_Y$$. We could define a topology on $$X$$, $$\tau_X$$, as follows

$\tau_x := f^{-1}(\tau_Y) = \{ U \subset X \; \vert \; \exists V \in Y \text{ s.t. } f^{-1}(V) = U\}$

This makes $$f: (X, \tau_X) \to (Y, \tau_Y)$$ continuous. It’s also the coarsest topology on $$X$$ such that $$f$$ is continuous. If $$X \subset Y$$ and $$f$$ is the inclusion map, then we call $$\tau_X$$ the subspace topology.

## Bijections and Homeomorphisms

If $$f$$ is a bijection, we have $$f^{-1}$$ well defined and $$(f^{-1})^{-1}=f$$. As such,

Theorem: Let $$f: (X, \tau_X) \to (Y, \tau_Y)$$ be a bijection. Then $$f$$ is continuous iff $$f^{-1}(\tau_Y) \subseteq \tau_X$$ iff $$\tau_Y \subseteq f(\tau_X)$$

It follows that

Theorem: Let $$f: (X, \tau_X) \to (Y, \tau_Y)$$ be a continuous bijection with a continuous inverse, then $$f(\tau_X) = \tau_Y$$

Proof: As $$f^{-1}: Y \to X$$ continuous, from the previous theorem we have $$(f^{-1})^{-1} (\tau_x) \subseteq \tau_Y$$ so $$f(\tau_X) \subseteq \tau_Y$$. Similarly as $$f$$ is continuous we have $$\tau_Y \subseteq f(\tau_X)$$. Result follows by double inclusion. $$\square$$

This suggests that homemorphisms preserve all topological properties.