Not necessarily true. If you have a particular bitstream that approximates a waveform, and you use a fourier transform to generate an approximation with sufficient accuracy that it reproduces the same bitstream, you have something lossless.

Also, even if you don't go that far, many lossless algorithms operate by first providing an approximation and then including the (usually small) deltas between the approximation and the actual bitstream.

The Fourier transform itself is an exact mathematical transformation with an exact inverse transform. There's nothing lossy about it.

Mathematically, you are correct. Practically you are not if you consider the transform source.

What do you mean? If the original source is analog and sampled below its Nyquist rate in the analog-to-digital conversion, the process is indeed irreversible. But that all happens before any transforms from the time domain to the frequency domain are in play, so it's a separate issue.

Beyond that, discrete Fourier and cosine transforms as usually implemented are not fully reversible because of loss in precision. A colleague of mine blogged about the issue in the context of Haar transforms a few years ago: http://cbloomrants.blogspot.com/2008/09/09-08-08-1.html. By decomposing an orthogonal transform into shears as explained by Charles, you can design reversible fixed-precision variants of the DCT like binDCT: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.41.8...

Sorry it is a separate issue - I should have been more clear.

The original point the OP made was a bad example. Unfortunately I made a poor attempt at explaining that.