How to compute the homology group of connected sum of Klein bottle and $RP^2$?
I want to use mayer vietories sequence,I know Klein bottle remove a disk is $S^1 \vee S^1$，and $RP^2$ remove a disk is $S^1 $,but i can't write the generator.
Homology group of connected sum of Klein bottle and $RP^2$
algebraic-topologyhomology-cohomologymayer-vietoris-sequence
Related Solutions
Consider a map $f\colon S^1 \to S^1 \vee S^1$ that maps upper half of a circle to the first summand in orientation preserving way, and lower half to the second circle in orientation reversing way. Also denote $i_1$ and $i_2$ inclusions of $S^1$ to $S^1 \vee S^1$ as two summands. Composition of $f$ with folding map $p\colon S^1\vee S^1 \to S^1$ (both summands in orientation-preserving way) is evidently homotopic to a constant map. Looking on $H_1$ we have $\mathbb{Z} \xrightarrow{f_*} \mathbb{Z}\oplus \mathbb{Z} \xrightarrow{p_*} \mathbb{Z}$.
$p_*$ is addition: $S^1 \xrightarrow{i_1} S^1\vee S^1 \xrightarrow{p} S^1$ is identity (similary for $i_2$). $i_1$ induces $\mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z}$, to make sure it is an inclusion of the first summand we may use naturality of exact sequences for a map $(S^1, *) \xrightarrow{i_1} (S^1\vee S^1, \mathrm{im}\, i_2)$. Now we know $p_*(1, 0) = 1$ and $p_*(0, 1) = 1$, by linearity $p_*$ is addition.
Thus $f_*$ maps $1\in H_1(S^1)\cong \mathbb{Z}$ to some pair $(x, -x)\in H_1(S^1\vee S^1)$. To find out what $x$ is, consider the composition $S^1 \xrightarrow{f} S^1 \vee S^1 \to S^1$ where the second map collapses second summand to a point (it induces projection on a second coordinate in $\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ too, e.g. by naturality of exact sequences of pairs). By construction this map is homotopic to identity. Similary for the second summand in $S^1\vee S^1$ we get that orientation-reversing map induces $-1$ on homology.
That's quite cumbersome to read, but actually very straightforward.
UPDATE
As mentioned in comments, here is proof reversing orientation induces identity on $H_0$. $* \to S^1$ and $*\to S^1 \to S^1$ (the second map is reversal of orientation) are equal as maps from a point, thus induce the same map on $H_0$. If we know that the inclusion of a base point in $S^1$ induces non-zero map on $H_0$, we're done, since automorphism of $H_0(S^1) \cong \mathbb{Z}$ fixing a non-zero element must be identity. To see this, use the naturality of long exact sequences of pairs for a map $(S^1, *) \to (*, *)$. The first commutative square (with non-relative $H_0$s) consists of three identity mapping of a point and a map induced from inclusion of a base point, and cannot commute unless $H_0(*) \to H_0(S^1)$ is an inclusion.
(Reply to a comment: nullhomotopic map induces zero on $H_1$ since it factors as map $X \to * \to Y$)
Also it seems to me that $X \hookrightarrow X \vee Y$ induces inclusion to the first summand in homology not due to functoriality properties of long exact sequences but this is just the way we identify $H_i(X\vee Y)$ with $H_i(X)\oplus H_i(Y)$. That is, we look at the pair $(X \vee Y, Y)$, write long exact sequence, then apply excision to get rid of relative homology and get $H_i(Y)$ instead. After that, we note that a lot of arrows in long exact sequence admit splitting and define the isomorphism $H_i(X \vee Y) \to H_i(X) \oplus H_i(Y)$ the way inclusion of $X$ in a wedge sum is inclusion on first summand. Thus some arguments by naturality above are not needed.
The Mayer-Vietoris sequence is defined as the long exact sequence
$$H_*(U \cap V) \to H_*(U) \oplus H_*(V) \to H_*(X)$$
where the first map is defined by $(i_*, j_*)$ and the second map is defined by $k_* - l_*$ where $i, j$ are the inclusion maps of $U \cap V$ inside $U$ and $V$ and $k, l$ are the inclusions of $U$ and $V$ inside $X$.
If $X$ is the Klein bottle, $\{U, V\}$ cover of it by the two mobius strips, then $U \cap V$ is homotopy equivalent to the boundary of the two strips. I have to thus compute the maps $i_*, j_*$ induced on homology where $i, j$ are inclusions of the boundary map of the mobius strips into the Klein bottle.
That's precisely where we need the wrapping description. If $M$ is a moebius strip then the inclusion map $\partial M \hookrightarrow M$ could be understood as follows: deformation retract $M$ to it's core circle, which generates $H_1(M)$. Under this deformation retract $\partial M$ maps to twice the core circle. So $i_* : H_1(\partial M) \to H_1(M)$ is precisely the multiplication by $2$ map.
Thus, $i_*$ is multiplication by $2$ whereas $j_*$ is multiplication by $-2$ due to orientation issues. Tat tells you $\alpha$ sends $1$ to $(-2, 2)$ once you identify $H_1(U \cap V)$ with $\Bbb Z$ and $H_1(U) \oplus H_1(V)$ with $\Bbb Z \oplus \Bbb Z$.
Best Answer
So you have constructed the following long exact sequence but are unsure what the map $f$ is.
$$ \begin{array}{cccccc} &0 &\to& 0 &\to& H_2(K\#\mathbb{R}P^2)\\ \to& H_1(S^1)\cong\mathbb{Z} &\stackrel f\to& H_1(S^1\vee S^1)\oplus H_1(S^1)\cong \mathbb{Z}^3 &\to& H_1(K\#\mathbb{R}P^2)\\\stackrel 0\to& \mathbb{Z}&\to& \mathbb{Z}\oplus \mathbb{Z}&\to& \mathbb{Z}\\\to&0 \end{array} $$
As the inclusion of a point in the intersection goes to a point in each piece, the map $\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$ is $1\mapsto (1,1)$ which is injective. Hence it has kernel isomorphic to $0$.
Clearly \begin{eqnarray*} H_2(K\#\mathbb{R}P^2)&=&{\rm ker} f,\\ H_1(K\#\mathbb{R}P^2)&=&{\rm coker} f,\end{eqnarray*} so once we know $f$ we have the homology groups.
From the following diagrams, we can see the generator $[t]$ of $H_1$ of the intersection, maps to twice the generator $[c]$ of $H_1(\mathbb{R}P^2 \backslash D^2)$ and twice one of the generators $[a]$ of $H_1(K \backslash D^2)$:
$$[t]\mapsto [a]-[b]+[a]-[b]=2[a],\qquad [t]\mapsto [c]+[c]=2[c]$$
Thus \begin{eqnarray*} H_2(K\#\mathbb{R}P^2)&=&{\rm ker} f=0,\\ H_1(K\#\mathbb{R}P^2)&=&{\rm coker} f=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z},\\ H_0(K\#\mathbb{R}P^2)&=&\mathbb{Z}. \end{eqnarray*}