Fourier transformation picture
How to show that Fourier transform (please find picture 1) is a bijective transformation?
For injective, I did the following. Is it right? $${F}_1(x)=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_1(x)e^{-i\omega x}dx}$$ $${F}_2(x)=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_2(x)e^{-i\omega x}dx}$$ Suppose that ${F}_1(x)={F}_2(x)$, then $$\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_1(x)e^{-i\omega x}dx}=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f_2(x)e^{-i\omega x}dx}$$ $$=>\int^{\infty}_{-\infty}{f_1(x)e^{-i\omega x}dx}-\int^{\infty}_{-\infty}{f_2(x)e^{-i\omega x}dx}=0$$ $$=>\int^{\infty}_{-\infty}({f_1(x)-f_2(x))e^{-i\omega x}dx}=0$$ $$=>f_1(x)=f_2(x)$$
How can we show that it is surjective?
Assuming that the functions involved are nice enough (which appears to be done implicitly in the attached text anyway), this follows from the Fourier inversion theorem, which provides an explicit inverse for the Fourier transformation.