We can resolve the forces acting on the cyclist separately into vertical and horizontal components. The diagram shows just the vertical components: the weight W, and three forces exerted by different parts of the bicycle on the cyclist: Fp from the pedals, Fs from the saddle, and Fh from the handlebars. Each of these are the equal-and-opposite reaction force to the forces exerted by the cyclist on the bicycle.
The cyclist is not accelerating vertically so there can be no net vertical force on the cyclist: W=Fp + Fs + Fh.
Let's assume Fh, the force exerted on the cyclist by the handlebars, is constant (it will be small anyway, so it won't matter much even if it's not strictly constant). And W is clearly constant.
Fs = [W-Fh] - Fp, where the bit in square brackets is constant.
Then, if Fp increases, Fs decreases. Or, as many people have already said, the saddle takes whatever part of the weight is not taken through the pedals.
For a cyclist working at 100 W, delivering that power output at 80 rpm through 170 mm cranks, Fp, averaged over a revolution, is about 10 kgf. It presumably actually varies from closer to zero at 12/6'o'clock to probably nearer double, say 20 kgf, at 3/9'o'clock. For an 80 kg rider, that means that the amount of the weight taken by the pedals is in the region of 10-20% body weight, and the amount of weight transmitted through the saddle reduced accordingly - significant but not dramatic.
But the force on the pedals increases as the power output increases and as the cadence drops. Outputting 400 W at 60 rpm gets you to the whole of the body weight taken through the pedals and none through the saddle - which makes sense because that's what we do when we climb hills out of the saddle. (Well, not the 400 W bit for some of us, obvs.
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