Calculating homology of a Klein bottle (Using only axioms)

Question

so let K be the Klein Bottle. I am perfectly aware how to calculate it for singular homology, using some properties of chains, and an explicit description of the differential. This is not my problem.

Let us take a (unreduced) homology theory and suppose that we have that $$h_n(S^1,\emptyset)=\mathbb{Z}$$ if $n=0,1$ and otherwise $0$. I want to calculate the homology of the Klein Bottle using only this fact, and that the functors $h_n$ satisfy the Eilenberg–Steenrod axioms. On page 109 of Switzer's book on Algebraic Topology, he shows how this can be done. The idea is to cut the klein bottle $K$ into two cylinders $A,B$ and that $A \cap B \cong S^1 \amalg S^1$ and use Mayer-Vietoris: $$\cdots \rightarrow h_n(A \cap B, \emptyset) \xrightarrow{\alpha} h_n(A,\emptyset) \oplus h_n(B, \emptyset) \xrightarrow{\beta} h_n(K, \emptyset) \xrightarrow{\Delta'} \cdots$$ and that $h_n(A\cap B, \emptyset) = \mathbb{Z} \oplus \mathbb{Z}$, $h_n(A,\emptyset) =\mathbb{Z}$, $h_n(B,\emptyset) = \mathbb{Z}$. One can show that the map $\beta$ is represented by a matrix $$\pmatrix{1 & 1 \\ 1 & v'_\ast}$$ where $v':S^1 \rightarrow S^1$ is the map reversing the orientation. Switzer then states that, $v_\ast:H^1(S^1,\emptyset) \rightarrow H^1(S^1,\emptyset)$ is $-1$ and $H^0(S^1,\emptyset) \rightarrow H^0(S^1,\emptyset)$ is $1$. Now here is my question (finally): How can we determine the sign of the map? It is easy to see that the map is either $\pm 1$, but the sign is more mysterious...

Answer 1

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.