# Concept of quotient map and quotient topology.

I am studying quotient topology.Different books define the concept differently but I think I have got the crux of the concept.I want to verify whether I have understood properly and whether it is sufficient for me.

Let $$X$$ and $$Y$$ be two topological spaces and $$f:X\to Y$$ be a surjection.Then $$f$$ is called a quotient map if $$U\subset Y$$ is open iff $$f^{-1}(U)\subset X$$ is open.Now suppose $$X$$ is a topological space and $$Y$$ is a set such that $$f:X\to Y$$ is a surjection.Then the topology on $$Y$$ that makes the map $$f$$ into a quotient map is the topology $$\mathcal T(f)=\{U\subset Y:f^{-1}(U)$$ is open in $$X\}$$.It is known as the quotient topology.Now if we have an equivalence relation $$\sim$$ on $$X$$ and $$X^*=X/\sim$$ be the quotient set,then together with the quotient topology $$X^*$$ is called the quotient space.Now $$\mathcal T(f)$$ on $$X^*$$ is the largest topology that makes the map $$f$$ continuous because if $$\tau$$ is another topology on $$X^*$$ such that $$f$$ is continuous then $$\tau\subset \mathcal T(f)$$.

Is this understanding fine and enough?

## 1 Answer

Let $$X$$ be a topological space, $$Y$$ be a set and $$f:X\to Y$$ be a map of sets. We define the final topology $$\mathscr{T}_f$$ on $$Y$$ relative to $$f$$ to be the largest topology on $$Y$$ making $$f$$ continuous; this topology is exactely defined by $$\mathscr{T}_f=\{U\subseteq Y : f^{-1}(U) \ is \ open \ in \ X \}$$ Now consider an equivalence relation $$\sim$$ in $$X$$ and $$\pi:X\to X/\sim$$ the quotient map. The quotient topology in $$X/\sim$$ is defined to be the final topology on $$X/\sim$$ relative to $$\pi$$. Now if $$X$$ and $$Y$$ are topological spaces and $$f:X\to Y$$ is a surjection and a quotient map i.e, $$U\subseteq Y$$ is open iff $$f^{-1}(U)\subseteq X$$ is open. Then we define in $$X$$ the relation $$\sim$$ by: $$x\sim x^\prime \Leftrightarrow f(x)=f(x^\prime)$$. It is easy to see that $$\sim$$ is an equivalence relation and that $$X/\sim$$ endowed with the quotient topology is homeomorphic to $$Y$$.