Traction force on tire

Or more basically, suppose you have some torque at the sprocket. Would you have to exert more or less rotating force at the tire to create that same torque at the sprocket? Remember that torque = force times radius.

 

Exactly. The farther you get from the center of rotation the less the torque. So the torque at the sprocket is not the same as the torque at the tire.

 

[tex]T_{tire} = T_{sprocket} - I\alpha[/tex]
Where [itex]I[/itex] is the inertia of the sprocket-axle-wheel-tire assembly and [itex]\alpha[/itex] is the rotational acceleration of the assembly.

If there is no acceleration ([itex]\alpha = 0[/itex]), then the tire torque is exactly the same as the sprocket torque. Otherwise, in typical assembly, the inertia should be small enough to consider that they are both practically the same as well (although, not necessarily a negligible difference).

 

The torque is absolutely the same. The force, however, is not, since as you pointed out earlier, torque is force multiplied by radius.

 

That is not always true, there can be losses between the sprocket and the wheel; friction at a bearing, losses through a CV joint, a dragging brake pad etc etc. Power still remains constant but not all of it makes it to the wheel:
P_sprocket ≠ P_Wheel

P_sprocket = P_Wheel + P_{bearing losses} P_{CV losses} + P_{any other losses}

Whether these losses are small enough to be ignored depends on your system.