In the following question $H^i_c(M;G)$ denotes the compact support cohomology, $M$ an $n-$oriented manifold and $G$ an abelian group.
The definition I have of Poincarè homomorphism $P_M : H^i_c(M;G) \longmapsto H_{n-i}(M;G)$ is the following:
The cap product operation $\mu_K$ defines an homomorphism $P_K: H^i(M,M\setminus K) \longmapsto H_{n-i}(M)$ defined by $P_K(\alpha) = \alpha \frown \mu_K$.
Taking two compacts $K \subseteq L$ and using the naturality of the cap product we have a commutative diagram
$$\require{AMScd}
\require{cancel}
\def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1}
\lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}}
\begin{CD}
&& H_{i}(M,M\setminus L)\\
& \diaguparrow{i} @VV\bullet \frown \mu_L V \\
H^{i}(M,M\setminus K) @>>\bullet \frown \mu_K> H_{n-i}(M)
\end{CD}$$
Taking the direct limit over all the compacts $K \subseteq M$ we have defined a map $P_M : H^i_c(M) \longmapsto H_{n-i}(M)$.
So as far as I understand the Poincaré map seems the map induced on the direct limit by this diagram.
In the consequences of the Theorem, I found in my notes the following observation:
$P_M(\alpha) = \alpha \frown \mu_M$ if $M$ is compact (I'm aware that if $M$ is compact than there exists a unique class in $H_n(M)$ which respect some properties of orientation).
Now my question is: how do we know that the Poincaré map, which is the map induced by taking the direct limit, is actually the homomorphism induced by taking the cap product with the fundamental class $\mu_M$ ?
Best Answer
If you are taking the colimit of a diagram with a terminal object, you can identify the colimit with the value of the diagram at the terminal object. In the case $M$ is compact, the terminal object is $H^i (M, \emptyset)=H^i(M)$. Since by definition your Poincare map on this object in the direct limit is capping with $\mu_M$, you are done.