$$
\newcommand{\H}[2]{\tilde{H}_{#1}(#2)}
\newcommand{\xto}[1]{\xrightarrow{#1}}
\newcommand{\im}{\operatorname{im}}
$$
I'm currently working on calculating the (reduced) singular homology of both the Klein bottle and projective plane, where I have stumbled upon a similar confusion in both examples.
For the Klein Bottle $K$, I used Mayer Vietoris with two Möbius bands $A,B$ whose intersection is also a Möbius band, as the picture shows:
Hence each of these is homotopic to $S^1$, and thus we get that $\tilde{H}_n(K) = 0$ for $n > 2$. For the rest of the groups, we have the following short exact sequence
$$
0 \to \H{2}{K} \xto{\partial} \H{1}{A \cap B} \xto{i} \H{1}{A} \oplus \H{1}{B} \xto{j} \H{1}{K} \to 0
$$
from which we have to calculate the remaining groups. The sequence is exact, and so both
$$
\H{2}{K} = \im\partial = \ker i,
$$
and
$$
\H{1}{K} = \im j = \frac{\H{1}{A} \oplus \H{1}{B}}{\ker j} = \frac{\H{1}{A} \oplus \H{1}{B}}{\im i}.
$$
Therefore to complete the calculation it is sufficient to "understand" $i$. If I recall correctly, the mapping is defined as the inclusion to each factor,
$$
\begin{align}
i : \H{1}{A &\cap B} \to \H{1}{A} \oplus \H{1}{B} \\
& \bar{x} \longmapsto (\bar{x},-\bar{x})
\end{align}
$$
I have convinced myself that since $\H{1}{A \cap B} \simeq \mathbb{Z}$, all cycles are homologous and so for example the group is generated by $\bar{c}$ the central circumference of the Möbius strip $A \cap B$. I can also intuitively see that when we look at $c$ in both $A$ and $B$, this is like going around these strips twice. Hence $i$ should look like $1 \mapsto (1,-2)$, when looking at the groups via the isomorphisms to $\mathbb{Z}$, which actually gives the desired result of $H_2(K) = 0$ and $H_1(K) = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. However, it is not clear to me how this 'homotopic intuition' matches the algebra, since $i$ as defined would seem to map $\bar{c}$ to generators of both $A$ and $B$, hence looking like $1 \mapsto (1,-1)$ which would in turn give the incorrect conclusion that $H_1(K) = \mathbb{Z}$.
Similarly, I've encountered the same issue with the (real) projective plane. In a similar approach, I defined $X = \mathbb{P}^2(\mathbb{R})$ as $D^2$ with antipodal identifications in the boundary, and then I have chosen the decomposition $A = X \setminus \{0\}$, $B = B_\varepsilon(0)$ from which to apply Mayer Vietoris. This again gives quite immeadiatly that $\H{n}{X} = 0$ for $n > 2$ and once again we get that
$$
0 \to \H{2}{X} \xto{\partial} \H{1}{A \cap B} \xto{i} \H{1}{A} \oplus \H{1}{B} \xto{j} \H{1}{X} \to 0
$$
from which we see that it is sufficient to characterize $i$, and I come across the same misunderstanding: in the 'homotopical context', it is clear to me that when embedding the circumference $\bar{c} \in \H{1}{A \cap B}$ into $\H{1}{A}$, it will be travelling twice the path of a generator, and this matches the fact that $\H{1}{X} = \mathbb{Z}/2\mathbb{Z}$ (i.e. $i$ is like $1 \mapsto (2,0)$). But I don't see how this coincides with the homological construction: when seeing $\bar{c}$ in $\H{1}{A}$, the latter is isomorphic to $\mathbb{Z}$ and so my intuition tells me that all curves are homologous, and moreover the map is defined as sending $\bar{c}$ to the formal combination $\bar{c} \in \H{1}{A \cap B}$ (or am I wrong here?) so I can't see where the $2$ comes from.
I apologize for the lengthy post, but I figured appropriate context was necessary in order for this to be a stand alone question.
Thoughts?
Best Answer
$\newcommand{\im}{\operatorname{im}}$Following your notation, let $\bar{c}\in H_1(A\cap B)$ be the homology class of the central circumference of the Möbius strip $A \cap B$, traversed from the bottom of the square to its top and let $\bar{a}\in H_1(A)$ be the homology class of central circumference of the (blue) Möbius strip $A$, traversed from the top of the square to its bottom. The following picture shows $\bar{c}$ in red and $\bar{a}$ in blue:
One may cut up this piece of the Klein bottle into four triangles as follows:
There are singular 2-simplices $\sigma_1,\sigma_2,\sigma_3,\sigma_4$ each mapping the standard 2-simplex homeomorphically into one of those four triangles and having their boundaries oriented as in the following diagram:
This shows that the boundary of the 2-chain $\sigma_1+\sigma_2+\sigma_3+\sigma_4$ is precisely $c+2a$ (note that all of the black edges cancel out with each other). Therefore, $\overline{c+2a}=0$ as an element of $H_1(A)$. A similar argument proves an analogous statement about $H_1(B)$.
Now, under the identifications $H_1(A\cap B)\simeq \Bbb{Z}$ taking $\bar{c}$ to $1$ and $H_1(A)\oplus H_1(B)\simeq \Bbb{Z}^2$ taking $\bar{a}$ and the analogous generator of $H_1(B)$ to $(1,0)$ and $(0,1)$ respectively, the map $i$ corresponds to the assignment $\Bbb{Z}\ni 1\mapsto (-2,-2)\in\Bbb{Z}^2$. This establishes the desired isomorphism $$H_1(K)\simeq \frac{H_1(A)\oplus H_1(B)}{\im i}\simeq \frac{\Bbb{Z}^2}{(-2,-2)\Bbb{Z}^2}\simeq \Bbb{Z}\oplus\Bbb{Z}/2\Bbb{Z}.$$