Oriented manifold. Connecting homomorphism.

From: L.P (mathjlp_at_yahoo.com)
Date: 03/09/05 

Date: 9 Mar 2005 02:42:02 -0800

Hello,

In an article where in the intro it says all manifolds are supposed to
be smooth, compact and orientable, a closed manifold M^n is defined to
be oriented if a generator for H_n(M^n) is chosen. So implicitly
connectedness is assumed as well, right ??

By Poincaré duality :

   H_n(M) = H^0(M) = Hom(H_0(M), Z) = Hom(Z+Z+...+Z, Z) = Z+Z+...+Z

with one summand for each (path) component of M.

By the way, could someone tell me whether I am missing something
really obvious when I can't immediately see why, to a given generator
\mu\in H_{n-1}(bd(W^n)) there is a class in H_n(W, bd(W)) mapping to
\mu under the connecting homomorphism, when W is compact, smooth and
orientable.

Thanks in advance.