Angular momentum without any apparent torque I have seen the bicycle wheel experiment which is the most basic demonstration of gyroscopic precession.
I have a doubt in that demonstration.

We give the wheel a spin, and when we let go, it starts revolving around the vertical axis instead of falling. I understand the reason to that very well.  
My doubt is that,
since the wheel is revolving around the vertical axis, there must be a component of angular momentum in the vertical direction.
However, I don't see any torque in the vertical direction.
So what increased the angular momentum of the wheel in the vertical direction starting from zero to a fixed value?
 A: 
My doubt is that,
  since the wheel is revolving around the vertical axis, there must be a component of angular momentum in the vertical direction.

Remember that the wheel is spinning i.e. the wheel is not a stationary disk that is rotating about the vertical axis. If you were to take into account the rotation of the wheel around its own axis as well as its rotation about the vertical axis, you would find that its angular momentum is changing consistent with $\boldsymbol \tau=\dot{\mathbf L}$, where $\boldsymbol\tau$ is the torque of the wheel's weight about the point where its axle touches the vertical axis, and $\mathbf L$ is the angular momentum about this same point.
I remember this video being a good "demystification" of this system.
A: Your observation is correct.

Before I start my answer let me use the picture above to define three axes:

*

*Roll axis - the gyroscope wheel spins around the roll axis.

*Pitch axis - motion of the red frame.

*Swivel axis - motion of the yellow frame.

To understand gyroscopic precession it is key to also recognize how motion of the pitch axis is playing a part.
As you point out: before and after the start of gyroscopic precession the angular momentum must be the same.
So: how does the spinning wheel acquire angular momentum around the swivel axis?
Let me take the case where you release the spinning wheel gingerly. You release it in such a way to that you avoid imparting angular momentum around the swivel axis, but you keep the wheel from bobbing up and down.
As you release the spinning wheel, it drops a little. In terms of angular momentum: the roll axis changes direction (pitches down). That change of direction of the roll axis (of the spinning wheel) accounts for the change of angular momentum around the swivel axis.
It's crucial to make the following distinction:

*

*the effect that starts the motion of gyroscopic precession.

*the effect that sustains the dynamic state of gyroscopic precession.

As we know, when you have a steady state of gyroscopic precession the force that is acting is not doing work. The following analogy is very commonly offered:

*

*A centripetal force that is sustaining circular motion isn't doing work. The centripetal force is necessary to sustain the circular motion.


*A torque that is sustaining gyroscopic precession isn't doing work. The torque is necessary to sustain the gyroscopic precession.
The shift from not-precessing to a state of gyroscopic precession is due to the wheel pitching down a little. Once the appropriate precession rate is reached any pitching motion ceases.
The importance of the wheel pitching down a little tends to be overlooked.
In demonstrations with for example a bicycle wheel the wheel is usually released rather roughly, which leaves the precessing wheel bobbing up and down (nutation). That tends to mask that the final pitch angle is always lower than the initial pitch angle
That is why your general observation is spot on.
This can be seen as follows:
Hypothetically, is it concievable that a wheel starts to precess instead of any pitching motion? No, that would violate the laws of motion. If the wheel would start to precess instead of any pitching you would have a violation of conservation of angular momentum.

Incidentally, the very same question that you raised was also raised, and answered, by Richard Feynman, in the Feynman Lectures;
Feynman lectures, chapter 20 Rotation in space

Some people like to say that when one exerts a torque on a gyroscope,
it turns and it precesses, and that the torque produces the
precession.
[...]
When the motion settles down, the axis of the gyro
is a little bit lower than it was at the start. Why? (These are the
more complicated details, but we bring them in because we do not want
the reader to get the idea that the gyroscope is an absolute miracle.
It is a wonderful thing, but it is not a miracle.) If we were holding
the axis absolutely horizontally, and suddenly let go, then the simple
precession equation would tell us that it precesses, that it goes
around in a horizontal plane. But that is impossible! Although we
neglected it before, it is true that the wheel has some moment of
inertia about the precession axis, and if it is moving about that
axis, even slowly, it has a weak angular momentum about the axis.
Where did it come from? If the pivots are perfect, there is no torque
about the vertical axis. How then does it get to precess if there is
no change in the angular momentum? The answer is that the cycloidal
motion of the end of the axis damps down to the average, steady motion
of the center of the equivalent rolling circle. That is, it settles
down a little bit low. Because it is low, the spin angular momentum
now has a small vertical component, which is exactly what is needed
for the precession. So you see it has to go down a little, in order to
go around. It has to yield a little bit to the gravity; by turning its
axis down a little bit, it maintains the rotation about the vertical
axis. That, then, is the way a gyroscope works.

This discussion by Feynman inspired Svilen Kostov and Daniel Hammer to perform a table-top experiment. The article describing their results is named after the key observation in Feynman's discussion 'It has to go down a little, in order to go around'
