Angle $\phi$ in astronomical coordinate transformations 
I was reading about the transformation between horizontal coordinates (Alt-Az) and equatorial coordinates (RA-Dec) from Fundamental Astronomy by H.Karttunen.
In the image, one can see the angle $\phi$.
My opinion is that $\phi$ remains constant and is equal to 23.5$^\circ$, where 23.5$^\circ$ is the tilt in the earth's axis of rotation from the geographic meridian. 
Is this true, or does angle $\phi$ change with latitude? If it does, can you explain how? 
 A: 
My opinion is that ϕ remains constant and is equal to 23.5°, where 23.5° is the tilt in the earth's axis of rotation from the geographic meridian. Is this true, or does angle ϕ change with latitude? If it does, can you explain how?

No, the obliquity of the ecliptic
($23.5°$, the tilt between the ecliptical plane and the equatorial plane,
traditionally noted as $\varepsilon$)
has nothing to do with the transformation between horizontal and equatorial coordinates.
The angle $\phi$ in your picture (between equator and zenith)
is just equal to the geographical latitude.
To convince yourself about this,
consider the easiest special cases first:


*

*Suppose you (the observer, in the center of the sphere)
are standing at the north pole (geographical latitude $\phi=90°$) of the earth.
The horizon and the equator of the sky are identical,
hence the tilt between horizon and sky-equator is $90°-\phi=0°$.
Also, the zenith and the north pole of the sky are identical,
hence the angle between $Z$ (zenith) and $P$ (north pole of sky) is $90°-\phi=0°$.

*Suppose you are standing at equator of the earth (geographical latitude $\phi=0°$).
Then the equator of the sky (from eastern point of horizon, through zenith,
to western point of horizon) is at right angle to the horizon,
hence the tilt between horizon and equator of the sky is $90°-\phi=90°$.
Now, the zenith and the north pole of the sky (at the northern point of the horizon)
are separated by an angle $90°-\phi=90°$.

