Editing Comparison of topologies (section)

== Properties ==

Let ''τ''<sub>1</sub> and ''τ''<sub>2</sub> be two topologies on a set ''X''. Then the following statements are equivalent:
* ''τ''<sub>1</sub> ⊆ ''τ''<sub>2</sub>
* the [[identity function|identity map]] id<sub>X</sub> : (''X'', ''τ''<sub>2</sub>) → (''X'', ''τ''<sub>1</sub>) is a [[continuous map (topology)|continuous map]].
* the identity map id<sub>X</sub> : (''X'', ''τ''<sub>1</sub>) → (''X'', ''τ''<sub>2</sub>) is a [[open map|strongly/relatively open map]].

(The identity map id<sub>X</sub> is [[surjective function|surjective]] and therefore it is strongly open if and only if it is relatively open.)

Two immediate corollaries of the above equivalent statements are
*A continuous map ''f'' : ''X'' → ''Y'' remains continuous if the topology on ''Y'' becomes ''coarser'' or the topology on ''X'' ''finer''.
*An open (resp. closed) map ''f'' : ''X'' → ''Y'' remains open (resp. closed) if the topology on ''Y'' becomes ''finer'' or the topology on ''X'' ''coarser''.

One can also compare topologies using [[neighborhood base]]s. Let ''τ''<sub>1</sub> and ''τ''<sub>2</sub> be two topologies on a set ''X'' and let ''B''<sub>''i''</sub>(''x'') be a local base for the topology ''τ''<sub>''i''</sub> at ''x'' ∈ ''X'' for ''i'' = 1,2. Then ''τ''<sub>1</sub> ⊆ ''τ''<sub>2</sub> if and only if for all ''x'' ∈ ''X'', each open set ''U''<sub>1</sub> in ''B''<sub>1</sub>(''x'') contains some open set ''U''<sub>2</sub> in ''B''<sub>2</sub>(''x''). Intuitively, this makes sense: a finer topology should have smaller neighborhoods.