Quotient Spaces:

Suppose that $X$ is a topological space and $f : X \longrightarrow Y$ is a surjective mapping, let us consider the various topologies on $Y$ with respect to which $f$ is continuous. Certainly the function $f$ would be continuous if $Y$ carries the trivial topology where the only open sets are $\emptyset$ and $Y$. The quotient topology on $Y$ is the strongest topology that makes $f$ continuous. More explicitly consider the family

$${\cal T} = \{ A \subset Y : f^{-1}(A)$$    is open in $$X \}. \eqno(4.1)$$

Since ${\cal T}$ is closed under arbitrary unions, finite intersections and contains $Y$ and the empty set, we conclude that ${\cal T}$ is a topology on $Y$ with respect to which $f$ is continuous. It is also clear that any strictly larger topology would render $f$ discontinuous.