Proposition

The quotient topology is indeed a topology.

Solution

Let $(X, \mathcal{T})$ be a topological space, $Y$ be any set, $q: X \rightarrow Y$ be a surjective map. Let $\mathcal{T}’ = \{ V \subset Y \mid q^{-1}(V) \in \mathcal{T} \}$.

We claim that $\mathcal{T}’$ is a topology.

• $q^{-1}(\emptyset) = \emptyset$, so $\emptyset \in \mathcal{T}’$.
• $q^{-1}(Y) = X$ because $q$ is surjective. Thus $Y \in \mathcal{T}’$.
• Let $\{ V_{\alpha} \} \subset \mathcal{T}’$. $q^{-1}(\bigcup_{\alpha} V_{\alpha}) = \bigcup_{\alpha} q^{-1}(V_{\alpha})$. Then each $q^{-1}(V_{\alpha}) \in \mathcal{T}$, so the union is in $\mathcal{T}$. Therefore, $\bigcup_{\alpha} V_{\alpha} \in \mathcal{T}’$.
• Let $V_{\alpha_1}, \cdots, V_{\alpha_n} \in \mathcal{T}’$. $q^{-1}(\bigcap_{i=1}^n V_{\alpha_i}) = \bigcap_{i=1}^n q^{-1}(V_{\alpha_i})$. Then each $q^{-1}(V_{\alpha_i}) \in \mathcal{T}$, so the finite intersection is in $\mathcal{T}$. Therefore, $\bigcap_{i=1}^n V_{\alpha_i} \in \mathcal{T}’$.

Therefore, $\mathcal{T}’$ is a topology.