$f^\to$ is our image operator, i.e. $f^\to(A_\gamma) = \{f(x) \in Y: x \in A_\gamma\}$
I seek to prove that $f^\to$ preserves joins, i.e. $\forall \{A_\gamma: \gamma \in \Gamma\} \subset \wp (X)$, for any set $X$
$$f^\to (\displaystyle\bigcup_{\gamma \in \Gamma} A_\gamma) = \displaystyle\bigcup_{\gamma \in \Gamma} f^\to (A_\gamma)$$
I also seek to do this in two cases: $\Gamma = \varnothing$ and $\Gamma \neq \varnothing$
Proof
Case 1: $\Gamma \neq \varnothing$
I start this attempt by considering the RHS, i.e. $\displaystyle\bigcup_{\gamma \in \Gamma} f^\to(A_\gamma)$
By definition, for $f: X \to Y$ and $A_\gamma \subset X,$
$$f^\to(A_\gamma) = \{f(x) \in Y: x \in A_\gamma\}$$
Then, rewriting $\displaystyle\bigcup_{\gamma \in \Gamma} f^\to(A_\gamma)$, we have equivalently
$$\displaystyle\bigcup_{\gamma \in \Gamma} f^\to(A_\gamma)= \displaystyle\bigcup_{\gamma \in \Gamma} \{f(x) \in Y: x \in A_\gamma\}$$
Now, since $\Gamma \neq \varnothing, \exists \gamma \in \Gamma$ and thus
$$\gamma \in \Gamma \Rightarrow \exists A_\gamma \in \{A_\gamma: \gamma \in \Gamma\} \Rightarrow \{A_\gamma: \gamma \in \Gamma\} \neq \varnothing$$
Thus, $\displaystyle\bigcup f^\to(A_\gamma) \neq \displaystyle\bigcup \varnothing = \varnothing$
But from here, my confidence wavers: How might I proceed to show that the RHS is equivalent to the LHS, i.e. that $f^\to$ preserves joins?
