Why is every map to an indiscrete space continuous? Show that if $Y$ is a topological space, then every map $f:Y \rightarrow X$ is continuous when $X$ has the indiscrete topology. Proof: Assume $X$ has the indiscrete topology, $T=\\{\varnothing,X\\}$. $f$ is continuous if $f^{-1}(V)$ is an open subset of $X$ whenever $V$ is an open subset of $Y$. Let $V$ be an open subset of $Y$. I dont know how to use this to show $f^{-1}(V)$ is an open subset of $X$.

You got confused about the definition of continuity.

If $f\colon Y\to X$ is continuous then the preimage of open subsets of $X$ is open in $Y$.

Since $X$ has the indiscrete topology, we only have two open subsets. Namely, $X$ and $\varnothing$.

The preimage of the empty set is of course empty, and therefore open in $Y$. If we look at $f^{-1}(X) = \\{y\in Y\mid f(y)\in X\\}=Y$, and of course that $Y$ is open in $Y$.

Thus, $f$ is continuous regardless to the topology given on $Y$ whenever $X$ is indiscrete.

**Exercise:** Suppose $f\colon X\to Y$ and $X$ has the discrete topology, prove that $f$ is continuous.