[proofplan] We prove that every subset of $Y$ is open. The [quotient topology](/page/Quotient%20Topology) is defined by testing openness after taking preimages under $q$. Since $X$ is discrete, the preimage of any subset of $Y$ is automatically open in $X$, so the quotient-topology criterion makes that subset open in $Y$. [/proofplan] [step:Show that an arbitrary subset of $Y$ is open in the quotient topology] Let $U \subset Y$ be arbitrary. The preimage of $U$ under the map $q: X \to Y$ is the subset \begin{align*} q^{-1}(U) = \{x \in X : q(x) \in U\} \end{align*} of $X$. Since $(X,\tau_X)$ is discrete, every subset of $X$ belongs to $\tau_X$; hence $q^{-1}(U) \in \tau_X$. By the definition of the quotient topology $\tau_Y$ induced by $q$, a subset $V \subset Y$ belongs to $\tau_Y$ exactly when $q^{-1}(V) \in \tau_X$. Applying this criterion to $V = U$, we obtain $U \in \tau_Y$. [guided] We need to prove that $(Y,\tau_Y)$ is discrete, which means that every subset of $Y$ is open in the topology $\tau_Y$. Therefore we begin with an arbitrary subset $U \subset Y$ and prove that $U \in \tau_Y$. The quotient topology induced by the map $q: X \to Y$ is defined by the following criterion: \begin{align*} V \in \tau_Y \iff q^{-1}(V) \in \tau_X \end{align*} for every subset $V \subset Y$. Thus, to show that our chosen subset $U$ is open in $Y$, it is enough to show that its preimage under $q$ is open in $X$. The preimage of $U$ is the subset \begin{align*} q^{-1}(U) = \{x \in X : q(x) \in U\} \end{align*} of $X$. Since $(X,\tau_X)$ is discrete, every subset of $X$ is open. In particular, this specific subset $q^{-1}(U)$ satisfies \begin{align*} q^{-1}(U) \in \tau_X. \end{align*} Now the quotient-topology criterion applies with $V = U$, and gives \begin{align*} U \in \tau_Y. \end{align*} So the arbitrary subset $U \subset Y$ is open in the quotient topology. [/guided] [/step] [step:Conclude that $Y$ is discrete] The subset $U \subset Y$ was arbitrary, and the previous step proves $U \in \tau_Y$ for every subset $U$ of $Y$. Therefore every subset of $Y$ is open in $\tau_Y$. This is exactly the definition of the [discrete topology](/page/Discrete%20Topology) on $Y$, so $(Y,\tau_Y)$ is discrete. [/step]