Let $(Y, \tau_Y)$ be a pseudo metric space, and let $\tau_Y$ be the associated topology. Let $\sim$ be the equivalence relation defined by $$x \sim y \text{ if } d_Y(x,y)=0,$$ for $x, y \in Y$. Set $X = Y/\sim$, and let $(X, d_X)$ be the quotient metric space. Consider the mapping $P: Y \rightarrow X$ defined by $x \mapsto [x]$. Let $\tau_D$ be the direct image of topology $\tau_Y$ on $X$ via the mapping $P$, and let $\tau_X$ be the topology of the metric space $(X, d_X)$. Show that $\tau_D = \tau_X$.
I'm not sure how to start on this problem. I think I need to prove that open sets of $\tau_D$ are contained in open sets of $\tau_X$ and vice versa. Open balls generate $\tau_X$ because it's a metric topology, but the problem is that I don't know what the open sets of $\tau_D$ are. Any assistance is appreciated.
I think you can prove it by showing $\tau_D$ and $\tau_X$ contain each other.
$\tau_D\subset \tau_X$. This is because each open ball in $\tau_Y$ is mapped onto an open ball in $\tau_ X$.
$\tau_X\subset \tau_D$. This is because $\tau_D$ is the quotient topology, and $P\colon (Y,\tau_Y)\to(X,\tau_X)$ is continuous, since $P^{-1}(B_X(x,\varepsilon))=\bigcup_{P(y)=x}B_Y(y,\varepsilon)$ for each $x\in X$ and $\varepsilon> 0$, where $B_X(x,\varepsilon)$ stands for the open ball in $X$ centered at $x$ with radius $\varepsilon$.