Show that an inverse image $F^{-1}$ is an isometric bijection, if $F$ is an isometric bijection.

Question

Let $X$ and $Y$ be normed spaces. In order to show the result, we would need to show that $\Vert F^{-1}(y)\Vert_X = \Vert y \Vert_Y$ for all $y \in Y$. My intuition tells me that bijectivity alone should suffice to prove this, as it implies that $$ F(F^{-1}(y)) = y \quad\text{and}\quad F^{-1}(F(x)) = x $$ for all $y \in Y$ and $x \in X$. Then taking the norm of $y = F(x) \in Y$ results in $$ \Vert y \Vert_Y = \Vert F(x) \Vert_Y = \Vert F(F^{-1}(y)) \Vert_Y = \Vert F^{-1}(y) \Vert_X $$ by the isometricity of $F$, where we have denoted $x = F^{-1}(y)$. Is this sufficient to conclude the proof?