Csdrhrt

$$
newcommand{H}[2]{tilde{H}_{#1}(#2)}
newcommand{xto}[1]{xrightarrow{#1}}
newcommand{im}{operatorname{im}}
$$

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

$$
H{2}{K} = impartial = ker i,
$$

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 $H_1(K) = mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$. 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 $H{1}{X} = mathbb{Z}/2mathbb{Z}$ (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?

Following your notation, let $bar{c}in H_1(Acap B)$ 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 $H_1(Acap B)simeq Bbb{Z}$ 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 $Bbb{Z}ni 1mapsto (-2,-2)inBbb{Z}^2$.
This establishes the desired isomorphism
$$H_1(K)simeq frac{H_1(A)oplus H_1(B)}{im i}simeq frac{Bbb{Z}^2}{(-2,-2)Bbb{Z}^2}simeq Bbb{Z}oplusBbb{Z}/2Bbb{Z}.$$