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I need to find a representation of $ [M \mathbin\sharp N] \in H_n(M \mathbin\sharp N) $ in terms of the fundamental classes $[M]$ and $[N]$. My idea is that $$ [M \mathbin\sharp N] = [M]+[N]$$

which is based on this MO question, whose answer is a little bit unclear to me because I never made use of triangulation for example.

To prevent some questions, The proof I gave for the fact that $M \mathbin\sharp N$ is orientable iff $M,N$ are, made use of the fact that removing an embedded euclidean nhbd doesn't change orientation behaviour. In fact I think that my question can be rephrased as a request for an homological proof of the orientation behaviour of the connected sum (even though there are lots of different ways to show this).

What I tried is to use the l.e.s of pair $(M \mathbin\sharp N , S^{n-1})$ together with M-V sequence of $M \vee N$ in order to "obtain" the plus, because by M-V we have an iso $$ H_n(M \vee N) \xleftarrow{incl_1 - incl_2} H_n(M) \oplus H_n(N) $$

and by l.e.s. pair we have a s.e.s. (some details can be found here)

$$\require{AMScd} \begin{CD} 0 @>>> H_n(M \mathbin\sharp N) @>>> H_n(M \mathbin\sharp N, S^{n-1}) @>>> H_{n-1}(S^{n-1}) @>>> 0 \\ @. @. @V{\cong}VV \\ @. @. \tilde{H}_n(M \mathbin\sharp N / S^{n-1}) \\ @. @. @V{\cong}VV \\ @. @. \tilde{H}_{n-1}(M \vee N) \end{CD}$$

But I'm unsure how to glue everything together, because the two inclusion with one minus sign would give to me exactly what i want, because I flip one orientation of one of the two manifolds. but these all are intuitive reasonings and I don't know how to formalise them properly.

$$ ++++ $$

following the comments below, it seems that a reasonable way to prove this is to show that $[M]+[N]$ has the characteristic property of the fundamental class. My problem now, (as I double-checked my computations) s that i dont0 have a valid proof for showing that $[M]$ and $[N]$ are elements of $H_n(M\sharp N)$, because trivially $M \not\subset M \sharp N$ obviously, so we need to find something that still represent $[M]$ i the homology group.

An idea was to use Poincaré Duality (see comments) to have a map $H_n(M) \to H_n(M \sharp N)$ but now the problem is that I don't know how to work with such image, because I don't have much control on how does the composition behave. At least, that's my problem right now

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    $\begingroup$ What about checking that $[M]+[N]$ gives you the local orientation generator on each point and concluding $[M\#N]=[M]+[N]$ from the uniqueness of the orientation class? See Hatcher Lemma 3.27. $\endgroup$ – archipelago May 26 '15 at 12:08
  • $\begingroup$ @archipelago I worked your idea a little bit, and it seems leading somewhere, but I cannot prove rigorously the characterising property in the case of points of $S^{n-1}$ (the one identified by the equivalence relation), could you elaborate a little bit this point? $\endgroup$ – Riccardo May 26 '15 at 15:39
  • $\begingroup$ In my humble opinion, the cleanest way is to construct the connected sum properly not to have issues on the gluing sphere, e.g. like in maths.ed.ac.uk/~aar/papers/kervmiln.pdf on the second half of page 505. $\endgroup$ – archipelago May 26 '15 at 16:14
  • $\begingroup$ @archipelago ok, I think the only missing proof in my reasoning is how to prove that $[M]+[N]$ is in fact an element of $H_n(M \sharp N)$. It is sufficient to check that $[M]$ and $[N]$ are in it, but I don't know how to show it, at the beginning I though about an "extending by zero" argument, but I feel that it's very sloppy and I cannot convince myself that's correct $\endgroup$ – Riccardo May 26 '15 at 20:27
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    $\begingroup$ You should probably define what $[M]$ and $[N]$ even mean in $H_*(M\#N)$. There are no natural inclusions $M$ or $N$ to $M\#N$. In fact there are not even say degree 1 maps from say a torus to the connect sum of two tori. There are in fact natural maps from $M\#N$ to $M$ or $N$ and if you define $\cap[M]$ and $\cap[N]$ using these maps and Poincare duality you should get the desired result. $\endgroup$ – PVAL-inactive May 26 '15 at 22:02

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