Question about the property 'one-to-one mapping' of the Fourier transform.

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I am a student studying on Electrical engineering. Recently, I am anxious about the fourier transform's mapping property during proving the relationship between bessel function and fourier transform of U-shape function.

My question is

If $G(f)$ exists such that $\mathcal{F}\left\{g(t)\right\}$ exists, where $\mathcal{F}$ represents the fourier transform, i.e., $$ G(f)=\mathcal{F}\left\{g(t)\right\} = \int_{-\infty}^{\infty} g(t) \exp\left(-j 2 \pi f t\right) dt,$$ then $g(t)$ exists and it is $\mathcal{F}^{-1}\left\{G(f)\right\}=\displaystyle\int_{-\infty}^{\infty}G(f)\exp\left(j2\pi ft\right)df$.

That is, when proving that $g(t)$ and $G(f)$ are pair in Fourier-transformation-sense, is it enough to prove only one direction either $\left(g\to G\right)$ or $(g\leftarrow G)$ without needing to prove both directions $(g\leftrightarrow G)$?