A description/proof of universal property of continuous embedding
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous embedding.
- The reader knows a definition of subspace topology.
Target Context
- The reader will have a description and a proof of the universal property of continuous embedding: any injection between topological spaces is a continuous embedding if and only if any additional map from any additional topological space into the domain of the original map is continuous if and only if the composition of the additional map before the original map is continuous.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any topological spaces, \(T_1, T_2\), and any injection, \(f_1: T_1 \rightarrow T_2\), \(f_1\) is a continuous embedding if and only if for any topological space, \(T_3\), and any map, \(f_2: T_3 \rightarrow T_1\), \(f_2\) is continuous if and only if \(f_1 \circ f_2: T_3 \rightarrow T_2\) is continuous.
2: Proof
Suppose that \(f_1\) is a continuous embedding. Suppose \(f_2\) is continuous. Then, \(f_1 \circ f_2\) is continuous as a compound of continuous maps. Suppose \(f_1 \circ f_2\) is continuous. Then, for any open set, \(U \subseteq T_1\), \(f_1 (U)\) is open on \(f_1 (T_1)\) while \(f_1 (U) = U' \cap f_1 (T_1)\) where \(U' \subseteq T_2\) is open on \(T_2\) by the definition of subspace topology. As \(f_1 \circ f_2\) is continuous, \((f_1 \circ f_2)^{-1} (U') = f_2^{-1} \circ f_1^{-1} (U')\) is open on \(T_3\). But \(f_1^{-1} (U') = f_1^{-1} (U' \cap f_1 (T_1)) = f_1^{-1} (f_1 (U))\), but \(f_1^{-1} (f_1 (U)) = U\), because \(f_1\) is bijective. So, \(f_2^{-1} (U)\) is open on \(T_3\), which means that \(f_2\) is continuous.
Suppose that for any topological space, \(T_3\), and any map, \(f_2: T_3 \rightarrow T_1\), \(f_2\) is continuous if and only if \(f_1 \circ f_2: T_3 \rightarrow T_2\) is continuous. Let us take \(T_3 = T_1\) and \(f_2: T_1 \rightarrow T_1\) as the identity map, continuous. So, \(f_1 \circ f_2 = f_1: T_1 \rightarrow T_2\) is continuous. As \(f_1\) is injective, \(f'_1: T_1 \rightarrow f_1 (T_1)\) is bijective, so, let us take \(T_3 := f_1 (T_1)\) where \(T_3\) is the subspace topological space and \(f_2 := {f'_1}^{-1}: f_1 (T_1) \rightarrow T_1\). \(f_1 \circ f_2: f_1 (T_1) \rightarrow T_2\) is an inclusion, so, continuous. So, \(f_2 = {f'_1}^{-1}\) is continuous, so, \(f'_1\) is a homeomorphism.
