TITLE: Atiyah duality without reference to an embedding
QUESTION [11 upvotes]: Atiyah duality is the equivalence $M/\partial M \simeq (M^{-T(M)})^\vee$, i.e. the Spanier-Whitehead dual of the space $M/\partial M$ is the Thom complex of the stable normal bundle of $M$. The theorem is proven by taking an appropriate embedding $i$ of $M$ into a sphere $S^d$, identifying the complement with the Thom complex of the normal bundle of the embedding, and constructing a duality map $M_+ \wedge M^{N(i)}\rightarrow S^d$ using the embedding.
In the case $M$ is a compact, framed manifold of dimension $n$, it is possible to describe a duality map $M_+ \wedge M_+ \rightarrow S^n$ without reference to an embedding by appealing to the fact that if we collapse everything outside a small ball around $p \in M$, we can use the framing to continuously identify it with $S^n$. This is the adjoint of a duality map $M_+ \wedge M_+ \rightarrow S^n$.
$\bf Question:$ Is it possible to create such a duality map for a framed manifold with nonempty boundary, namely one that does not invoke an embedding?

REPLY [9 votes]: Here is another short construction which is much simpler and just takes a few lines.

Let $M$ be a closed $n$-manifold. Consider the diagonal $M \to M \times M$. It is an embedding. Take its Pontryagin-Thom construction to get a map
$$
M_+ \wedge M_+ \to M^\tau
$$
(we have identified a tubular neighborhood of the diagonal with the total space of the tangent bundle).

If $M$ is (stably) framed, then $M^\tau \simeq M_+ \wedge S^n$ (stably). Then we have the map $M_+ \wedge S^n \to S^n$ induced by smashing $M_+ \to \text{pt}_+$ with $S^n$.

The composition
$$
M_+ \wedge M_+ \to M^\tau \simeq M_+ \wedge S^n \to S^n
$$
is what you want: it's a duality map.


This can be seen on the level of homology (but it is enough to check a map is a duality map on the level of homology).

If $M$ is compact with non-empty boundary $\partial M$, then there is a map of pairs
$$
(M\times M, M\times \partial M) \to (M\times M,M\times M - U)
$$ where $U$ is a tubular neighborhood of the diagonal.
To get this map, one might choose a collar neighborhood $C$ of $\partial M$ and thereafter identify $M$ with $M-C$.  Then
$M-C $ and $\partial M$ are disjoint.

The map of pairs determines a map of quotients
$$
M_+ \wedge M/\partial M \to M^\tau \, ,
$$
and one may then proceed as in the empty boundary case, assuming that $M$ is framed, to obtain a map
$$
M_+ \wedge M/\partial M \to S^n
$$
which will be an $S$-duality.