# Proof that the quotient topology is a topology

my book gives the following definition of quotient topology: let $f: X\to Y$ be a surjective mapping from a space $X$ onto a set $Y$. The quotient topology on $y$ with respect to $f$ is the family

$$\mathcal{U}_{f} = \lbrace U \vert f^{-1}(U) \text{ is open in X }\rbrace$$

When it comes to proving that the so-defined quotient topology is a topology, the book says that obviously $Y \in \mathcal{U}_{f}$, but I wonder why? Imagine there exists $U \subset Y$ such that $f^{-1}(U)$ is not open in $X$. What guarantees that $f^{-1}(Y)$ is open in $X$? I've tried to figure it out on my own to no avail, and when looking up the definition of quotient topology on the Internet, I found only definitions based on equivalence classes—which makes more sense, and probably that way it would be easier to see why a quotient topology is a topology, but I want to understand what I'm failing to see in this definition. Thanks in advance!

• Observe that $f^{-1}(Y)=X$ and secondly that $X$ is open in $X$. These observations allow you to conclude $Y\in\mathcal U_f$. – drhab Jul 11 '16 at 12:47
• I thought of that, but $f$ is only a surjection, so it could be $f^{-1}(Y)\subset X$ with a strict inclusion. Right? – Nicola Jul 11 '16 at 13:24
• No, that is not right. $f^{-1}(Y):=\{x\in X\mid f(x)\in Y\}$ and for every $x\in X$ it is true that $f(x)\in Y$, so every $x\in X$ is an element of $f^{-1}(Y)$. – drhab Jul 11 '16 at 13:42
• I don't understand what assumption says that every $x \in X$ is such that $f(x) \in Y$. That is, unless we're implicitly assuming that, if not all $x \in X$ are such, we restrict ourselves to all $x \in X$ that are such, and relabel the newly-obtained set $X$. – Nicola Jul 11 '16 at 14:28
• If $f:X\to Y$ is defined as a function then by definition $f(x)\in Y$ for every $x\in X$. In that context the set $Y$ is by definition the codomain of $f$. – drhab Jul 11 '16 at 14:31

For any map $f:X \to Y$, $f^{-1}(Y) = X$. So in particular $Y$ is open in the quotient topology.
Look: $f^{-1}(im(f)) = X$ and $im(f)\subset Y$. Hence $f^{-1} (Y)=X$ which is open in $X$.