# Calculating homology of a Klein bottle (Using only axioms)

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...

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.

• If $f\colon (X, A) \to (Y, B)$ is a mapping of pairs, we have two long exact sequences. For every group in the first long exact sequence $f$ induces map to the corresponding group in the second long exact sequences. Naturality here means that these "vertical" maps commute with "horizontal" ones, i.e. with maps included in long exact sequences. – Dmitry Aug 25 '13 at 20:08
• OK - would you mind expanding details on how the latter map induces -1 on homology? Excuse me for asking so much of you! – Tedar Aug 25 '13 at 20:10
• The composition of $f$ and projection to the first component is homotopic to identity map. By the reasoning in the asnwer we know that on homology the induced map is $1 \in H_1(S^1) \mapsto (x, -x) \mapsto x$. Then $x = 1$. Now look at the composition of $f$ and projection on the second summand: it is homotopic to orientation reversing automorphism and on homology it is $1\mapsto (x, -x) \mapsto -x$, that is, $-1$. – Dmitry Aug 25 '13 at 20:16
• This answer is essentially a proof that Hurewicz map from $\pi_k(X) \to H_k(X)$ is a homomorphism of groups. It works for all dimensions, not only for $\pi_1$. – Dmitry Aug 25 '13 at 20:35
• Maps $* \to S^1$ and $* \to S^1 \to S^1$ (here second map is the reversal of orientation) are equal, hence induce the same map on $H_0$. As long as the inclusion of base point induces non-zero map on $H_0$, we're done. To see it's not zero, use naturality of long exact sequences for pairs $(S^1, *) \to (*, *)$: it shows the induced map on $H_0$ is inclusion. I've asked related question earlier. Now I think it's way better to prove that singular homology satisfies the axioms and use it. – Dmitry Aug 25 '13 at 21:00