Homology computation I was computing the homology of a space obtained from 2 n-spheres $S_1^n,S_2^n$ by identifying them along their equatorial $(n-1)$-spheres. I had a doubt but I figured it out while typing, so I'm posting it anyway in case it is useful for someone.
I'm using two different methods to do it:


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*Using Mayer Vietoris: Let $U,V$ consist of the disjoint union of one sphere and an open neighborhood of the equator of the second one. In the quotient, $U\cap V$ retracts to the identified equatorial (n-1)-sphere and both $U,V$ retract to $S_1^n,S^n_2$ respectively, so the relevant portion of the Mayer-Vietoris sequence reads $$0 \rightarrow H_n(S^n_1)\oplus H_n(S_2^n) \rightarrow H_n(X) \rightarrow H_{n-1}(S^{n-1}) \rightarrow 0$$ so counting ranks we must have $H_n(X) = \mathbb{Z}^3$.

*Using cellular homology: we can construct our space by attaching 4 n-cells $A_1,A_2,B_1,B_2$ to an (n-1)-cell $\gamma$ so our cellular complex reads $$0 \rightarrow \mathbb{Z}^4 \stackrel{d_n}{\rightarrow} \mathbb{Z} \rightarrow 0 \rightarrow \cdots \rightarrow \mathbb{Z} \rightarrow 0$$ where $d_n(A_1)=d_n(B_1)=\gamma$ and $d_n(A_2)=d_n(B_2)=-\gamma$ and we get $H_n(X)=\ker d_n\simeq \mathbb{Z}^3$.
All other homology groups are clearly trivial except for $H_0(X)\simeq \mathbb{Z}$ since $X$ is path-connected.
 A: A nice way to find this answer 'intuitively' is to note that $X$ is homotopy equivalent $\mathbb{R}^{n+1}\setminus\{a,b,c\}$ for three distinct points $a,b,c$ in $(n+1)$-space. This is easiest to think about in the $n=1$ and maybe $n=2$ case but it generalises easily enough to all $n$. Now, the $i$th homology group $H_i$ is deigned in some sense to 'count' $i$-dimensional holes where you can think of an $i$-dimensional hole as being a part of the space which can be enclosed by a $i$-sphere in such a way that the sphere is not contractible in the space (or equivalently the enclusion of the sphere can not be extended to the inclusion of an entire disc).
We have $3$ holes in our space $X$ (up to homotopy at least) which can be enclosed each by an $n$-sphere $S^n$, and so $H_i(X)$ should be isomorphic to $\mathbb{Z}^3$. (As I said this is only an intuitive approach and isn't particularly rigorous for many hand-wavey reasons - this certainly wouldn't work for any space that can't be described as a subspace of $\mathbb{R}^n$ given by removing certain nice complementary subspaces)
A more rigorous, and perhaps computationally more efficient method compared to your two perfectly fine proofs, would be to note that $X$ is also homotopy equivalent to $S^n\vee S^n\vee S^n$, the wedge of three sphere whose reduced homology is then just given by the three-fold sum of the reduced homology of a single $n$-sphere.
