- TOPOLOGY AND GEOMETRY OF HYPERBOLIC 3-MANIFOLDS isi
7ri (E^2 ) Iker (i*) must be isomorphic to either 0, z, or z x Z.^2 By the classification
of abelian groups, ker ( i*) is isomorphic to Z provided ker ( i*) =f. ni (E).
On the other hand, by Corollary 31.15, a nontrivial element of ker (i*) is rep-
resented by a simple closed curve in E. Since this element must be primitive, it
generates ker (i*) ~ Z. 0
For the statement and proof of the next result, all homology and cohomology
coefficients are JR.
LEMMA 31.20 (Induced map on Hi for the boundary inclusion). Let M^3 be a
compact orientable 3 -manifold with boundary E^2 = 8M, which may be disconnected.
Let i : E Y M be the inclusion map. The rank of
i* : Hi (E) ---+ Hi (M)
is equal to half the dimension of Hi (E).
PROOF. Recall the following two basic facts from linear algebra. If A : !Rn ---+
!Rm is a linear map and if A* : !Rm ---+ !Rn denotes its dual, then
(31.7) dim(image(A)) = dim(image(A*)) = rank(A)
and
(31.8) dim (ker (A))+ dim (image (A)) = n.
We have the short exact (co )homology sequence
(31.9)
Hi (M) ~ Hi (E),
where D is the isomorphism from Lefschetz duality (this requires E = 8M) and
where i is the dual of i. Thus, by (31.7), we have
dim (image (j)) =dim (image (i)) =dim (image (i)).
Now the exactness of the sequence in (31.9) says
dim (image (j)) =dim (ker (i*)),
so that
dim (image (i)) =dim (ker (i)).
Finally, by (31.8), we have
dim (ker (i)) +dim (image (i)) =dim (Hi (E)),
so that dim (ker (i*)) =~dim (Hi (E)).^3 0
We have the following standard results from geometric topology, which we shall
use in the next subsection.
(^2) That is , we have ruled out the possibilities llm, Zn x Zm, and Z x Zn.