proving that a map $f : A \rightarrow B$ is a quotient map Suppose $A$ and $B$ are topological spaces such that $f : A \rightarrow B$ is a continuous surjective map. Assume that $\forall$ open set $U$ of $A$ its image is open. Then $f$ is a quotient map. 
The proof of this does not seem to bad, but I am still a little unsure. Usually when I think a proof is "not to bad" I start second guessing myself because I feel like it needs more when it doesn't. Is the proof of this as straight forward as I think it is or does it take a little more work? Maybe someone could show me their version of how to prove this. 
 A: I do think it's pretty straightforward. You need to check that $V$ is open in $B$ if and only if $f^{-1}(V)$ is open in $A$. The "if" is the openness condition together with surjectivity, and the "only if" is the definition of continuity. 
A: I'll show that if $f$ is a continuous surjective open map then it satisfies the universal property of a quotient map and so is itself a quotient map.
Let $C$ be a topological space and let $g\colon B\to C$ be a function. We need to show that $g$ is continuous if and only if $g\circ f$ is continuous.
Suppose that $g\circ f$ is continuous. Let $U$ be open in $C$ then because $g\circ f$ is continuous, we must have $V=(g\circ f)^{-1}(U)$ is an open subset of $A$ and so  $$f(V)=f((g\circ f)^{-1}(U))=f(f^{-1}(g^{-1}(U)))=g^{-1}(U)$$ is open because $f$ is an open map. It follows that $g$ is continuous by definition of continuity.
If we suppose that $g$ is continuous, then because $f$ is continuous we find that $g\circ f$ is also continuous because composition of continuous maps is continuous. It follows that $f$ satisfies the necessary universal property of a quotient map, as required.
