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?



Atiyah duality without reference to an embedding
Here is another short construction which is much simpler and just takes a few lines.


1. 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).
2. 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 $S^n$.
3. 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).


4. 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.


