If X is path connected, you can demonstrate base point invariance by drawing a path from b to any other base point in X, and building a correspondence. Review the earlier proof, and note that the constructed homotopies keep Sn-1 within X at all times. Thus the proof extends to the relative homotopy of Y/X.
If X is simply connected then πn(Y/X) corresponds to the unanchored homotopy classes of Dn/Sn-1 in Y/X. Apply the earlier proof.
Given an image of Sn-1 in F, follow the paths to get a homotopy on Sn-1 in Y, starting with an image in X, and ending at B. Homotopy means continuity, so let's have a look. Let U be an open set in Y, and let c be a point on our sphere, where c maps to a path that winds up in U at time t. In fact the path lies in U throughout time (t-ε, t+ε). Let K be the closed interval [t-ε/2,t+ε/2]. The paths that map K into U form an open set in Y~, which pulls back to an open set in the sphere, which contains c. Cross this with the open interval (t-ε/2,t+ε/2), and the result lies in U. This proves continuity, and the homotopy is valid.
For ease of illustration, let me proceed with D2/S1. Cross a circle with the unit interval I and get the wall of a can. Squeeze the top together to make a cone, which is homeomorphic to a disk. (In general, Sn-1 crossed with I, and closed at the top via a quotient space, is Dn.) After you have closed up the top, draw a line from the base to the top of the cone, corresponding to the base point a of the circle. Collapse this line down to a point via another quotient space. The result is still homeomorphic to a disk, with a at the boundary. Map this into Y using the aforementioned homotopy. Remember that the final circle of the homotopy maps to b, so there is no trouble collapsing this circle to a point. Also, a maps to the base point of Y~, which is the constant loop at b. So there is no trouble collapsing the line at a down to a point. We have a map from the disk into Y, with the boundary mapping to X, and a mapping to b.
This process is completely reversible. Draw a line from the boundary of the disk to a, and map this into Y, to find a path in F. Let c map to a path satisfying a KU constraint. Each point of K becomes a circle in our disk, tangent to a. These look like the rings of a shell, closing in towards the point along the edge. K becomes a collection of these rings, forming a closed set, and when the path from c to a slices through these rings, the intersection maps into U in Y. The preimage of U is an open set, actually a union of base open sets. We only care about the open sets that cover our chord intersect K. This is compact in the disk, hence finitely many open sets will do. Let r be the smallest radius of these open balls. Now the path can pivot at a, left or right, through any distance less than r, and the image still satisfies our KU constraint. Note that c sweeps out an open set on the boundary, hence the preimage of our open set in F is open, and the map is continuous. There is a one to one correspondence between the images of D2/S1 into Y/X, and the images of S1 into F.
A homotopy in either world becomes a homotopy in the other. Map one image to the other, as described above, for each time t. Of course you need to prove the resulting homotopy is continuous, but it is essentially the same proof as above. Cross the circle with the unit square, instead of the unit interval. The second dimension facilitates the homotopy. I'll leave the details to you. Therefore the homotopy classes correspond.
This shows that πn(Y/X) is a group for n > 1, and an abelian group for n > 2. Use the group structure of πn-1(F).
πn(F) → πn(Y~) → πn(Y)
The middle group is homotopy equivalent to X. The paths retract back to their starting points in X, and drag the image of the sphere along with. Thus we can replace the middle group with π(X).
πn(F) → πn(X) → πn(Y)
The second arrow is projection, which takes you to the end of the path. But paths have been retracted back to points, which start and end at the same place. So projection has become inclusion. Each sphere in X becomes a sphere in Y.
The paths of F also retract back to X, whence we only care about the image of the sphere at time 0. This makes both arrows inclusions.
As shown above, π(F) is isomorphic to π(Y/X). Turn a ball into paths, from the surface to a base point, then restrict to the start of these paths. This leaves the image of the spherical boundary of the ball. When F is viewed as Y/X, the first arrow, formerly known as inclusion, becomes the restriction of the image of the ball in Y to the image of the sphere in X.
πn+1(Y/X) → πn(X) → πn(Y) →
The third arrow, which takes you from one row down to the next, is the boundary map that turns a sphere into a homotopy on a lower sphere, lifts this homotopy, and takes the final image. Since the homotopy starts at b, it defines a disk in Y/X, or paths in F.
Rearrange the columns, to look more like X into Y with quotient Y/X. This has the added benefit of having the same dimension across each row.
πn(X) → πn(Y) → πn(Y/X) →
Generalizing, g could be any function taking X into Y, and we can construct the hurewicz map as before. This gives another long exact sequence in which the first arrow is induced by g, rather than a simple inclusion.
Since Y is simply connected, π1(Y) = 0. Similarly, any path in Y whose endpoints lie in X can shrink to a point in X. So the only nontrivial group on the bottom row is π1(X), which is Z.
The second row starts with 0, so we have a short exact sequence comprising the next two groups and Z from the bottom row. The epimorphism onto Z is the restriction of the map to the equator. In other words, the degree of the boundary of the disk, as it wraps around the equator, becomes the homotopy class of the circle around the circle on the bottom row. Conversely, given any degree d, one can wrap the disk around the northern hemisphere d times, thus wrapping the boundary around the equator d times. This looks like a reverse homomorphism from Z back to π2(Y/X), and we're just about ready to call the sequence split exact, but we need to prove this map is consistent with the structure of the relative homology, as a group. Remember that the group structure is inherited from the corresponding group π1(F), where F is the fiber. Turn the disk into a homotopy on the circle, starting at the base point b and winding up (no pun intended) in the equator X. Adding loops in X adds degrees, so the same holds for π(Y/X). The group acts as expected, the map from Z into π2(Y/X) is a reverse group homomorphism, and the sequence is split exact. If we can find the kernel, we have characterized π2(Y/X).
Map the disk into Y, so that the boundary has degree 0 around the equator. Stretchh and shrink the boundary of the disk, and the anulus adjacent to the boundary, until the boundary maps to b. Since the entire boundary maps to b, we really have a map from the sphere into the sphere. So we have proved that π2(S2) is the same as the homotopically distinct maps from the sphere into the sphere, with b mapping to b. That's basically the definition, so we really haven't made much progress. This gives you an idea of the difficulty of this topic. All that machinery, and we're back where we started.
When evaluating π3(Y/X), or anything higher for that matter, the boundary is a 2 sphere that maps into the circle. A homotopy shrinks this to a point, and the same homotopy extends to the disk. Now the boundary of the disk maps to a point, and we're back to mapping the 3 sphere into the 2 sphere. This demonstrates the equivalence between the second and third columns for n > 2.