I understand that, for $f \in S(\mathbb{R})$ (the Schwartz space) the transform \begin{equation} \tag1 Tf(\xi) = (2\pi)^{-\frac{1}{2}} \int_\mathbb{R} e^{i\xi x}f(x) \,dx \end{equation}
defines a left inverse for the Fourier Transform \begin{equation} f(x) \mapsto \mathcal{F}(\xi) = \hat{f}(\xi) = (2\pi)^{-\frac{1}{2}}\int_\mathbb{R} e^{-i\xi x} f(x) \, dx \end{equation}
Now I need to show that it is a right inverse, or (equivalently) that $\mathcal{F}$ is surjective.
Here is where I am struggeling to proceed. The proof that I am currently reading states that surjectivity follows from the fact that $(1)$ maps $f(x)$ to $\hat{f}(-\xi)$, but why does this mean that $\mathcal{F}$ is surjective?
Many thanks for your help!