Solution
= Solution
The map is <continuous> when the <preimage> $f^{-1}(U)$ is <open> in $X$ for every <open set> $U$ in $Y$. Equivalently the <preimage> of every <closed set> in $Y$ is closed in $X$, since preimages commute with complementation.
= Solution
The map is <continuous> when the <preimage> $f^{-1}(U)$ is <open> in $X$ for every <open set> $U$ in $Y$. Equivalently the <preimage> of every <closed set> in $Y$ is closed in $X$, since preimages commute with complementation.