If someone could look over this answer and let me know if I need to change anything it would be much appreciated!
I have posted the question along with my solution!
If someone could look over this answer and let me know if I need to change anything it would be much appreciated!
I have posted the question along with my solution!
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The argument is essentially correct: $f^{-1}[A] = X\setminus f^{-1}[Y\setminus A]$ is closed when $A$ is closed and $f$ continuous. And the same can be said with $A$ open and $f$ preserving inverse images of closed sets.