Let $f:X\rightarrow Y$ be a bijective function, and let $f^{-1}:Y\rightarrow X$ be its inverse. Let $V$ be any subset of $Y$. Prove that the forward image of $V$ under $f^{-1}$ is the same set as the inverse image of $V$ under $f$.
MY ATTEMPT
By definition, the inverse image of $V$ under $f$ is given by the set \begin{align*} f^{-1}(V) = \{x\in X \mid f(x)\in V\} \end{align*}
On the other hand, the forward image of $V$ under $f^{-1}$ is given by \begin{align*} f^{-1}(V) = \{f^{-1}(y)\in X\mid y \in V\} \end{align*}
Thus we have to prove that $\{x\in X\mid f(x)\in V\} = \{f^{-1}(y)\in X\mid y\in V\}$.
Let us consider the inclusion $(\subseteq)$ first.
If $a\in\{x\in X\mid f(x)\in V\}$, then $a\in X$ and $f(a) = v \in V\subseteq Y$.
Since $f$ is bijective, $a = f^{-1}(v)\in X$.
Hence $a\in\{f^{-1}(y)\in X\mid y\in V\}$.
Conversely, let us prove the inclusion $(\supseteq)$.
If $a\in\{f^{-1}(y)\in X\mid y\in V\}$, $a = f^{-1}(v)\in X$ and $v\in V$.
Once again, since $f$ is bijective, we conclude that $v = f(a)\in V\subseteq Y$.
Thus $a\in\{x\in X \mid f(x)\in V\}$.
Could someone double-check my reasoning?