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 $X$.
 A: To expand on Asaf's comment:
$f:X \rightarrow Y$ is continuous if $f^{-1}(O)$ is continuous for all open sets $O$ in $Y$. As $Y$ has the trivial topology, the only open sets are $\emptyset$ and $Y$. So to show that $f$ is continuous you need to show that $f^{-1}(\emptyset)$ and $f^{-1}(Y)$ are open, i.e. are in the topology of $X$.
A collection of sets is per definition a topology if it contains the entire space $X$ and $\emptyset$.  $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(Y) = X$ are therefore both open and so $f$ is continuous. 
A: 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.
