Mapping to Indiscrete Space is Continuous

Theorem
Let $T_1 = \left({S_1, \tau_1}\right)$ be any topological space.

Let $T_2 = \left({S_2, \tau_2}\right)$ be the indiscrete topological space on $S_2$.

Let $\phi: S_1 \to S_2$ be a mapping.

Then $\phi$ is continuous.

Proof
From the definition of continuous:
 * $U \in \tau_2 \implies \phi^{-1} \left({U}\right) \in \tau_1$

The only elements of $\tau_2$ are $S_2$ and $\varnothing$, from which:


 * $\phi^{-1} \left({S_2}\right) = S_1 \in \tau_1$


 * $\phi^{-1} \left({\varnothing}\right) = \varnothing \in \tau_1$