A (somewhat) conceptual proof that the boundary of a fundamental class of a manifold with boundary goes to a fundamental class? In this set-up, let M be a compact n-dimensional manifold with boundary $\partial M \neq \emptyset$. 
Assume that M is orientable, and that $[M] \in H_n(M,\partial M;R)$ is the fundamental class of M. If you haven't seen orientability / fundamental classes for manifolds with boundary before, the definition is the same, you just consider it for $M - \partial M$ instead. Now,one can consider the image of $[M]$ under the boundary map $\partial: H_n(M, \partial M;R) \rightarrow H_{n-1}(\partial M ;R)$ . I want to show that $\partial [M]$ is also a fundamental class of $\partial M$, and preferably in a manner that generalies to other homology theories, so a proof just using the Eilenberg-Steenrod axioms together with maybe something else would be nice. I can't seem to find one, so any help would be welcome!
 A: By the definition, an orientation on a manifold $M$ is a class $[M]\in h(M,\partial M)$ such that for any (interior) point its image under $h(M,M\setminus U_\epsilon(x))$ is a generator of $h(M,M\setminus U)\cong \tilde h(S^n)$ (considered as an $h(pt)$-module). Let's write, say, $[M]\bigr|_U$ for this element.
OK, let's move to the proof. Consider a small neighborhood $U$ of a point in $\partial M$. We need to show that $\partial$ maps $[M]\bigr|_U$ to $[\partial M]\bigr|_{U\cap\partial M}$. So everything reduces to the statement for $M=D^n$, $\partial M=S^{n-1}$ — that is, to well-known suspension isomorphism.
Maybe all this is clearer on a commutative diagram ($V:=U\cap\partial M$):
$$\require{AMScd}\begin{CD}
[M]\in h(M,\partial M) @>{\partial}>> h(\partial M);\\
@VVV  @VVV \\
[M]\bigr|_{U\setminus V}\in h(M,M\setminus (U\setminus V)) @. h(\partial M,\partial M\setminus V);\\
@|  @| \\
[D^n]\in h(D^n,S^{n-1})=\tilde h(S^n) @>{\partial}>>\tilde h(S^{n-1})\ni [S^{n-1}].
\end{CD}$$
(Summary: being an orientation is a local condition, so it can be changed locally — i.e. essentially just for $M=D^n$.)

(As Tedar points out) all this is explained in May's Concise Course in Algebraic Topology (see the proof of  «Proposition. An R-orientation of M determines an R-orientation of ∂M» on p. 169 + the proof of the next one) — the proof is essentially the same, but of course May's writing is better.
A: Here is an alternative way to check this. I don't know if it works for generalized homology theories, but I came here wanting to know the answer to your question for homology, and eventually learned this answer from Bredon's book (chapter 6. proposition 9.1). I think this is fairly conceptual, since it follows from some easy homological algebra applied to a naturally lying around LES of a triple. The hard vanishing statements involved in it are relatively easy consequences of an important theorem relating local homology and compactly supported sections of the orientation sheaf.
Anyway, I didn't really understand the other answer, but I think this is sufficiently different.
The idea of the proof is as follows: Suppose that $A$ is a component of $\partial M$, with $M$ an n-dimensional manifold. Let $B = \partial M \setminus A$. Then we consider the LES of the triple $(M, A \cup B = \partial M, B)$. Inside of it there is the exact sequence:
$H_n(M, B) \to H_n(M, \partial M) \to_{\partial} H_{n-1}(\partial M, B) \to H_{n-1}(M, B)$.
We use the following computations (proven below):


*

*$H_n(M,B; R) = 0$

*$H_{n-1}(\partial M, B) = \tilde{H}_{n-1} (A) \cong \mathbb{Z}$

*$H_n(M, \partial M) \cong \mathbb{Z}$

*$H_{n-1}(M,B)$ is torsion free.
To conclude that we have an exact sequence:
$0 \to \mathbb{Z} \to_{\partial} \mathbb{Z} \to \text{torsion free}$. Hence the map labeled with $\partial$ must be surjective.
Repeating this for each component of the boundary shows that the boundary of the fundamnetal class in $H_n(M, \partial M)$ is the fundamental class of $\partial M$.

Proof of the computations:
The main tool in these computations is the following theorem, which you can find (in a weaker form) as 3.27 in Hatcher.
Theorem 7.8 In Bredon's Topology and Geometry states that for closed $A$ in $X$, the map $J_A : H_n(M, M \setminus A ; G) \to \Gamma_c(A, \mathscr{O} \otimes G)$ is an isomorphism. $\mathscr{O}$ is the orientation sheaf. Here $J$ is the map that sends a homology class in $H_n(M, M \setminus A)$ to $H_n(M, M \setminus x ; G)$ for $a \in A$. $\Gamma_c$ refers to compactly supported global sections.
1.
Let $V$ be an open collar neighborhood of $B$.
Given this, we can make the following computations: $H_n(M,B; R) \cong_{\text{deformation retraction}} H_n(M, V) \cong_{\text{excise the boundary}} H_n(int(M), int(M) \cap V) \to_J \Gamma_c(int(M) \setminus V, \mathscr{O} \otimes G) = 0$.
(The latter group is 0, since $int(M) \setminus V$ is connected and non-compact... hence there are no compactly supported nonzero sections of a local system on it.)


*$\partial M$ is a closed orientable manifold, so this follows from Theorem 3.26 in Hatcher.

*By taking a collar neighborhood $V$ of $\partial M$, we get $H_n(M, \partial M) \cong H_n(M,V)
 \cong H_n (int(M), V \cap int(M)) \cong H_n (int(M) | N)$, with $N$ the complement of the collar neighoborhood in the interior. Next by the cited theorem, $H_n(int(M) |N ) \to \Gamma_c(N, \mathscr{O})$ is an isomorphism. Finally, by assumption $int(M)$ is orientable, and if we restrict a section of the  orientation sheaf to (compact $N$) , we get that $\Gamma_c(N, \mathscr{O})$ is nonzero.

*From universal coefficients $H_n(M,B; \mathbb{Q} / \mathbb{Z}) = H_n(M,B) \otimes \mathbb{Q} / \mathbb{Z} \oplus Torsion(H_{n-1}(M,B))$. By 1, this implies that $Torsion(H_{n-1}(M,B)) = 0 $. (To deduce this statement from universal coefficients, tensor $0 \to T(G) \to G \to G / T(G) \to 0$ with $Q / Z$, and tensor $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q} / \mathbb{Z} \to 0$ with $TG$. Here $T$means the torsion subgroup of an abelian group. The isomoprhisms that come out of the resulting LES for $\otimes$ will show that $Tor^1_{\mathbb{Z}}(H_{n-1}, \mathbb{Q} / \mathbb{Z}) = T(H_{n-1}(X,A))$. Which is a neat algebra lemma -- the first derived functor Tor against $\mathbb{Q}/\mathbb{Z}$ is exactly the torsion subgroup functor.)
