Let $S = \{1,2\} = T$ and define $f: S \to T$ by $f(1) = 1 = f(2)$. Let $B = \{1\}$ and let $C = \{2\}$. Find $f[B \cap C]$ and $f[B] \cap f[C]$.
I do not fully understand this, so obviously for the first one. $B \cap C$ is an empty set since they have nothing in common between the two sets, so would the function just result in an empty set? I am not sure how to compute the second one. Thank you.
You are right about the first part: $B\cap C=\emptyset$, and $f[\emptyset]=\emptyset$ regardless of what $f$ is.
Re: the second part, recall what "$f[X]$" means: it's $\{f(x): x\in X\}$. So $$f[B]=\{f(x): x\in B\}=\{f(1)\}$$ and $$f[C]=\{f(x): x\in C\}=\{f(2)\}.$$ Now:
What is $f(1)$?
What is $f(2)$?
Now you've found $f[B]$ and $f[C]$; what's their intersection?