Let $X$ be a topological space and let $\pi :X\rightarrow Y$ be a surjective map. Let Y be equipped with the quotient topology determined by $\pi$, obtained by declaring a subset $U\subseteq Y$ to be open if and only if $\pi^{-1}(U)$ is open in $X$.
If $K\subseteq Y$ is such that $\pi^{-1}(K)$ is closed in $X$, then there exists an open subset $V\subseteq X$ such that $\pi^{-1}(K)=X\setminus V$.
How can it be deduced that $K$ is closed for the quotient topology on $Y$? My problem is that from the definition of the quotient topology on $Y$ we cannot say that there exists some open subset $U\subseteq X$ such that $V=\pi^{-1}(U)$ (but if this holds then we could conclude from the surjectivity of $\pi$).
If $\pi^{-1}(K)$ is closed in $X$ then $X\setminus\pi^{-1}(K)=\pi^{-1}(Y\setminus K)$ is open, and so $Y\setminus K$ is open in $Y$. Thus $K$ is closed.