Confused by what this function image set is.

Question

I am given that a function is continuous, i.e.

$$f: X \to Y,$$ and I'm to prove that

$$f: X \to f[Y]$$ is also continuous when $f[Y]$ has the subspace topology. From my understanding of the notation, $f[X] = \{f(x):x \in X\}$. So, how am I suppose to know if $f$ can map co-domain values, i.e. $f(y)$ where $y \in Y$ unless $Y \subseteq X$? Am I missing something or is this a typo of $f[X] \subseteq Y$?

Answer 1

Yes, it should be f(X) as a subspace of Y. Quite trivial to prove, they just want to see that you have learned that an open set on a subspace is an intersection of open in ambient space with that subspace... Then you use a simple property of inverse image commuting with intersections, and voila, you will be done

Answer 2

I presume this is a typo, and that instead of $f[Y]$, it should be $f[X]$. To prove it, note that $f^{-1}(U) = f^{-1}(U \cap f[X])$ for all $U \subseteq X$.