Higher Order π Groups, Boundary Map

Boundary Map

Assume Y~ is serre over Y. Let b be a base point for Y. Let F be the fiber of b. Let b~ be a point in F, which acts as the base point for Y~. Since F embeds in Y~, there is an inclusion map from π_n(F) into π_n(Y~). By the projection of Y~ onto Y, there is a map from π_n(Y~) into π_n(Y). Finally, and this is nonobvious, there is a boundary map from π_n(Y) into π_n-1(F).

Let V be the image of Sⁿ in Y, such that the base point of the sphere maps to b. Express Sⁿ as the suspension of S^n-1. Move from one side of the suspension to the other, and V becomes a homotopy on S^n-1, that starts and ends with the trivial map from S^n-1 onto b. Every sphere is a finite cw complex, and Y~ is serre over Y. Let b~ be the lift at time 0, and lift the entire homotopy. The final image, at t = 1, becomes the boundary map. This necessarily lies in F, since the image downstairs lies in b.

We must show this is a well defined map on homotopy classes. Assume we have two homotopic images V and W of Sⁿ in Y. Each image of Sⁿ, from V to W, is itself a homotopy of S^n-1, so that the composite is a two dimensional homotopy on S^n-1 with two time coordinates. In other words, the n-1 sphere is crossed with the unit square, and this space is mapped continuously into Y. The bottom of the square represents V and the top represents W. The left side maps to b. In other words, we assume the homotopy maps the base point of the sphere into b at all times. Define a new homotopy with one time coordinate that sweeps out the area of the square, starting at the left, bottom, and top, and ending on the right side. At time 0 this homotopy has a lift. Lift the homotopy and look at the time 1 projection, the right side of the square. This becomes a homotopy between the two boundary spheres, completely in F, since the right side of the square maps to b.

This addresses another problem that you might have missed. Y~ is not a covering space, and the lift is not unique. What if we choose a different lift for V? Remember that V is trivially homotopic to V. Fill the square with the homotopy on homotopies, and give one lift to V at the bottom, and a different lift to V at the top. The left lifts to b~ as usual. This creates a valid lift for S^n-1 cross a path, and we can lift the homotopy, showing the boundary images are indeed homotopic. Put this all together and boundary is a well defined operator on π groups.

Homomorphism

For n > 1, the boundary map is a group homomorphism. Remember that V+W in Y is homotopic to the image of a single sphere. A great circle, running through the base point, is compressed down to the base point, squeezing one sphere into two. Then V and W are applied to the two smaller spheres. The implied homotopy on S^n-1 is, at every time t, V_t+W_t, joined at b in Y. This carries through all the way to t = 1. The lift, at every time t, is the image of two n-1 spheres joined at b~ in Y~. Evaluate this from t = 0 to 1, and the boundary of V+W is equal to, or at least homotopic to, the boundary of V + the boundary of W.

Note that the constant sphere at b lifts to a constant sphere at b~, which has a constant boundary at b~. In other words, 0 in π_n(Y) maps to 0 in π_n-1(F).

At The Bottom

Let n = 1, and consider a loop at b. Lift this to a path from b~ to another point in F. If Y~ is path connected, then draw a path from b~ to any point c in F. Project this down to Y, and find a loop in π₁(Y) that has b~,c as its boundary. All of F is accessible.

Long Exact

The map from inclusion to projection to boundary is long exact. The image of each group homomorphism is the kernel of the next. Exactness at the middle is easy. A sphere in F, included into Y~, and mapped down to Y, winds up in b, and a sphere whose projection is homotopic to b in Y is homotopic to a sphere in F; just lift the homotopy up to Y~.

Moving to π(F), the first column, a sphere in the boundary is homotopic to b~ in Y~ by construction. Thus any boundary becomes 0 in π(Y~). Conversely, assume a sphere becomes homotopic to b~ when it is allowed to move freely through Y~. Project this homotopy down to Y, and run it in reverse, and our original sphere in F is the boundary of a higher sphere from below.

Finally, look at π(Y), the third column. Project a sphere U in Y~ down to a sphere V in Y. When computing the boundary of V, let U act as the lift. The homotopy downstairs starts at b, sweeps an n-1 sphere around the surface of V, and winds up back at b. The lifted homotopy upstairs starts at b~, sweeps an n-1 sphere around the surface of U, and winds up at b~. Thus b~ is the boundary, which is 0 in π(F).

Conversely, let the boundary of V be homotopic to b~ in F. If U is a lift of V, then U implies a homotopy from b~ to a sphere in F, which is in turn homotopic to b~. Concatenate these homotopies to build a new homotopy from b~ back to b~. This is the image of a sphere. Project this down to Y and find a homotopy that runs through V for t = [0,½], then sits at b for t in [½,1]. This is a reparameterization of V. In other words, the new sphere is homotopic to V, and V is in the image of π(Y~). That completes the proof.