Quotient topology

Note: Motivation for quotient topology may be useful

(Unknown grade)

This page is currently being refactored (along with many others)

Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.

Definition of the Quotient topology

$$\begin{xy} \xymatrix{ X \ar[r]^p \ar[dr]_f & Q \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy}$$

(OLD)Definition of Quotient topology

If

$$(X,\mathcal{J})$$ is a topological space, $$A$$ is a set, and $$p:(X,\mathcal{J})\rightarrow A$$ is a surjective map then there exists exactly one topology $$\mathcal{J}_Q$$ relative to which $$p$$ is a quotient map. This is the quotient topology induced by $$p$$

[Expand]

The quotient topology is actually a topology

---

---

TODO: Easy enough

---

Quotient map

Let $(X,\mathcal{J})$ and $(Y,\mathcal{K})$ be topological spaces and let $p:X\rightarrow Y$ be a surjective map.

$p$ is a quotient map if we have

$$U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}$$

That is to say

$$\mathcal{K}=\{V\in\mathcal{P}(Y)|p^{-1}(V)\in\mathcal{J}\}$$

Also known as:

• Identification map

Stronger than continuity

If we had $\mathcal{K}=\{\emptyset,Y\}$ then $p$ is automatically continuous (as it is surjective), the point is that $\mathcal{K}$ is the largest topology we can define on $Y$ such that $p$ is continuous

Theorems

[Expand]

Theorem: The quotient topology, $\mathcal{Q}$ is the largest topology such that the quotient map, $p$, is continuous. That is to say any other topology such on $Y$ such that $p$ is continuous is contained in the quotient topology

---

For a map $p:X\rightarrow Y$ where $(X,\mathcal{J})$ is a Topological space we will show that the topology on $Y$ given by:

• $$\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}$$

is the largest topology on $Y$ we can have such that $p$ is continuous

Proof method: suppose there's a larger topology, reach a contradiction.

Suppose that $\mathcal{K}$ is any topology on $Y$ and that $p:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$ is continuous.

Suppose that $\mathcal{K}\ne\mathcal{Q}$

Let $V\in\mathcal{K}$ such that $V\notin \mathcal{Q}$

By continuity of $p$, $p^{-1}(V)\in\mathcal{J}$

This contradicts that $V\notin\mathcal{Q}$ as $\mathcal{Q}$ contains all subsets of $Y$ whose inverse image (preimage) is open in $X$

Thus any topology on $Y$ where $p$ is continuous is contained in the quotient topology

This theorem hints at the Characteristic property of the quotient topology

Quotient space

Given a Topological space $(X,\mathcal{J})$ and an Equivalence relation $\sim$, then the map:

$$q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})$$ with $$q:p\mapsto[p]$$ (which is a quotient map) is continuous (as above)

The topological space $(\tfrac{X}{\sim},\mathcal{Q})$ is the quotient space where $\mathcal{Q}$ is the topology induced by the quotient

Also known as:

• Identification space

See also

• Open and closed maps