Proving two Topologies to be equal

Question

Let $(X,\mathcal T)$ and $(Y,\mathcal V)$ be topological spaces, let $f$ be a function that maps $X$ onto $Y$, and let $\mathcal U$ be the quotient topology on $Y$ induced by $f$. Proved that if $f$ is continuous and closed then $\mathcal U= \mathcal V$.

All I can deduce with the information given in the question is that $f$ is quotient map.

Can you please tell how should I proceed with this problem. I am not good in problems related to quotient topology and I am still learning.Just telling the main elements of proof will be sufficient.I will fill details myself.

Thanks!

Answer 1

Since $\mathcal U = \{U \in \mathcal P(Y) : f^{-1}(U) \in \mathcal T\}$, we see that $\mathcal U$ is the biggest topology on $Y$ that makes $f$ continuous, that is, if $\mathcal T'$ is a topology on $Y$ such that $f : (X,\mathcal T) \to (Y,\mathcal T')$ is continuous, then $\mathcal T' \subseteq \mathcal U$. Why? Well, if $U \in \mathcal T'$, then $f^{-1}(U) \in \mathcal T$ (this is just the continuity of $f$), and this means precisely that $U \in \mathcal U$, so, $\mathcal T' \subseteq \mathcal U$.

Once observed this, if $f : (X,\mathcal T) \to (Y,\mathcal V)$ is continuous, $\mathcal V \subseteq \mathcal U$. On the other hand, if $U \in \mathcal U$, then $f^{-1}(U) \in \mathcal T$ (by definition) and $f(X \setminus f^{-1}(U))$ is closed in $(Y,\mathcal V)$ (since $f$ is closed). But, $$f(X \setminus f^{-1}(U)) = f(f^{-1}(Y \setminus U)) \stackrel{!}= Y \setminus U$$ meaning that $U \in \mathcal V$, and so, $\mathcal U \subseteq \mathcal V$ (what happens in $(!)$?).

By the way, if you know that those conditions implies that $f$ is a quotient map, then $\mathcal U = \mathcal V$ by merely definition of quotient map, in other words, this problem is more elementary.

Answer 2

Let $C$ be closed for $\mathcal{U}$, the quotient topology. This implies that $f^{-1}[C]$ is closed for $\mathcal{T}$, because of the definition of the quotient topology. As we assume that $f: (X,\mathcal{U}) \to (Y, \mathcal V)$ is closed (and continuous) we see that $C= f[f^{-1}[C]$ (equality holds as $f$ is onto) is closed for $\mathcal{V}$ as the image under $f$ of $f^{-1}[C]$. So $\mathcal{U} \subseteq \mathcal{V}$.

Let $C$ be closed for $\mathcal{V}$. As we assume that $f: (X,\mathcal{U}) \to (Y, \mathcal V)$ is continuous, we know that $f^{-1}[C]$ is closed for $\mathcal T$, and so as again by definition of the qutient topology $\mathcal U$ we conclude that $C$ is closed for $\mathcal U$. So $\mathcal{V} \subseteq \mathcal{U}$, and we have equality of topologies.