Hatcher Exercise 2.1.17

We compute in each of the following scenarios: Throughout, we will reference the long exact sequence:

(a): , is a finite set of points. Clearly, for , we have , so it must be the case that . Consider the LES in low dimensions:

Filling this in with known values, we have:

First, we must have .

is actually the inclusion . Then it must be the map sending every generator of to since every generator of (one of the points in ) is homologous to the single generator of when considered in the (connected) space . This map is clearly surjective. Since it’s a map between free modules over a PID, and is rank 1, we can conclude that is free of rank , i.e. .

Knowing the details of , we can find the rest of the unknown groups from repated applications of exactness and the first isomorphism theorem. . Likewise we find .

(b): and is a finite set of points. This is very similar to the last problem. The first part proceeds identically so that we have for . Now, the LES is low dimensions (starting at ) is:

Again, we must have . The map is the same one from the previous part. As such, follows by the same argument.

Since is surrounded by free groups in the exact sequence, it must be free (citation needed). Thus, by the same sort of reasoning as before, we conclude that its rank must be the sum of that of and , which is isomorphic to since the map to the left of is injective. That is, must be free of rank and we get .

Note: We can do part (a) and (b) in an easier way – though arguably against the spirit of the problem. Since is a manifold and is just a finite set of points, they certainly form a good pair (appropriately small open neighborhoods around each point deformation retract onto the point). Then we have . In either case, is homotopy equivalent to the wedge sum of with copies of . The best way to see this is inductively. Take , and identify two of the points . This is homotopy equivalent to attaching an external arc outside of connected to and . Then, take some path in connecting and , and contract it – we’re left with a wedge sum of and . For each remaining point , add another arc and contract along the path from the wedge base point to to add another copy of to the wedge sum. Continuing in this way, we end up with . Then, since this is a wedge sum of manifolds, each pair, consisting of each summand with the wedge base point, is a good pair and we can apply Corollary 2.25 in Hatcher to directly compute the homology based on the homology of the summands. When , we get:

And when :

Which agrees with what we get from considering the long exact sequence.

(c) and is the copy of used in performing the connect sum (the loop dividing into two connected components).

Again is a deformation of a small annalus around , so is a good pair. Then, it’s easy to see that is homotopy equivalent to . Then, , which is easy to compute again using Corollary 2.25. We have:

With for .

(d) is , and is a copy of subdividing one of the individual tori into two connected components.

The same conditions apply, so we have . Considering for one of the tori, observe that this space is homotopy equivalent to the sphere with two points identified (contract to a point, and then split that point into two), which is homotopy equivalent to . So then the space is homotopy equivalent to , but , so we have . So, we have:

And for .