Topology MOC

Quotient topology

The quotient topology is the canonical way of defining a topology on a Algebraic quotient, as defined by an Equivalence relation or projection. Let (𝑋,T𝑋) be a topological space, and πœ‹ :𝑋 ↠𝑆1 be a surjective function. The quotient topology on 𝑆 ≅𝑋/πœ‹ is the finest topology for which πœ‹ is continuous.2 #m/def/topology

Tπœ‹={π‘ˆβŠ†π‘†:πœ‹βˆ’1π‘ˆβˆˆT𝑋}

Further characterisations

Universal property

For every topological space (𝑍,T𝑍) and 𝑓 :𝑆 →𝑍, then 𝑓 is continuous iff π‘“πœ‹ :𝑋 →𝑍. #m/thm/topology

Proof

First we will prove that the quotient topology as characterised above satisfies the universal property. Let (𝑋,T𝑋) be a topological space, πœ‹ :𝑋 ↠𝑆 be a surjective function, and 𝑆 be endowed with the quotient topology Tπœ‹. Let (𝑍,T𝑍) be some topological space, and let 𝑓 :𝑆 →𝑍 be a function. If 𝑓 is continuous, then so is the composition π‘“πœ‹ of continuous functions. Now suppose π‘“πœ‹ :𝑋 →𝑍 is continuous, and let π‘ˆ ∈T𝑍. Then (π‘“πœ‹)βˆ’1π‘ˆ =πœ‹βˆ’1π‘“βˆ’1π‘ˆ ∈T𝑋 whence π‘“βˆ’1 ∈T𝑆. Thus 𝑓 is continuous. Therefore 𝑓 is continuous iff π‘“πœ‹ is continuous.

Now let Tβ€² be a topology on 𝑆 satisfying the universal property. In particular, let (𝑍,T𝑍) =(𝑆,Tπœ‹) and 𝑓 =id𝑆 :𝑠 ↦𝑠. Then since π‘“πœ‹ =πœ‹ is continuous so is 𝑓, wherefore Tβ€² is finer than Tπœ‹ Now let (𝑍,T𝑍) =(𝑆,Tβ€²) and 𝑓 =idπ‘Œ. Since idπ‘Œ is continuous, so too is idπ‘Œβ‘πœ‹ =πœ‹. But Tπœ‹ is the finest topology for which πœ‹ is continuous, so Tπœ‹ =Tβ€².

Further terminology

Properties

Spaces constructed as quotients


#state/tidy | #lang/en | #SemBr

Footnotes

  1. where 𝑆 is often constructed as the fibres of πœ‹, which is precisely the Algebraic quotientient]] 𝑋/ ∼ ↩

  2. 2020, Topology: A categorical approach, pp. 28–29 ↩