Klein bottle, description of isomorphism between cellular/simplicial homology, abelianization map from $\pi_1$ to $H_1$? Denote the Klein bottle by $K$. I can compute that$$\pi_1(K) \cong \langle a, b \mid ab = ba^{-1}\rangle, \quad H_0(K) = \mathbb{Z}, \quad H_1(K) = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}, \quad H_2(K) = 0.$$
In what follows, I have already searched online for an answer, but could not find anything satisfactory, for reasons I will explain.
Question 1. What is a description of the isomorphism between the cellular and simplicial homology specifically here?
There is the following theorem in Hatcher, but I am interested in a description of the isomorphism between the cellular and simplicial homology in the specific case of a Klein bottle, and I would like to avoid using this theorem and see such a description from more basic principles.

Theorem 2.35. $H_n^{\text{CW}}(X) \approx H_n(X)$.

Question 2. What is a description of the abelianization map from $\pi_1$ to $H_1$ specifically here?
We can use the Hurewicz theorem or stuff from Section 2.A of Hatcher, but again, I would like to avoid using those and see such a description from more basic principles.
 A: *

*The map $\pi_1 (X,x_0) \to H_1 (X)$ is the obvious one: a loop is mapped to the corresponding $1$-simplex $\sigma\colon \Delta^1 \to X$ with $\partial_0 (\sigma) = \partial_1 (\sigma) = x_0$. In your case, the loops $a$ and $b$ correspond to two $1$-simplices $\sigma_a$ and $\sigma_b$ and the relation $ab = ba^{-1}$ corresponds to $\sigma_a + \sigma_b = \sigma_b - \sigma_a$, i.e. $2\,\sigma_a = 0$. The result is $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, as expected.


*In Theorem 2.35 that you quote $H_n (X)$ denotes the singular homology, not simplicial homology. If you are reading Hatcher, then the isomorphism $H_n^{CW} (X) \cong H_n (X)$ holds simply because in the book, the complex for calculating $H_n^{CW} (X)$ is by definition built from the relative singular homology groups $H_n (X^n, X^{n-1})$, with appropriate differentials defined from the long exact sequences of pairs $(X^n, X^{n-1})$ — see the lemma and the commutative diagram before Theorem 2.35.
The groups $H_n (X^n, X^{n-1})$ are free, generated by $n$-cells, so in your case $H_1 (X^1,X^0)$ will correspond to the free abelian group generated by the two $1$-cells $a$ and $b$, and $H_2 (X^2,X^1)$ and $H_0 (X^0)$ will be free abelian groups on one generator, corresponding to the one $2$-cell and the one $0$-cell respectively. The differentials may be written down from the attaching maps (see the discussion after Theorem 2.35):
$$0 \to \mathbb{Z} \xrightarrow{d_2\colon 1 \mapsto 2a} \mathbb{Z} \left<a\right>\oplus \mathbb{Z} \left<b\right> \xrightarrow{d_1} \mathbb{Z} \to 0$$
Here $d_1 = 0$ (because the space is connected and we know that $H_0 \cong \mathbb{Z}$), so you get $H_1 \cong \frac{\mathbb{Z} \left<a\right>\oplus \mathbb{Z} \left<b\right>}{\mathbb{Z} \left<2a\right>} \cong \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. As $d_2$ is injective, you recover the fact that $H_2 = 0$.


*As for simplicial homology, whenever you have a triangulation, it is a particular case of cellular decomposition, and the corresponding cellular chain complex will be literally the same thing as the simplicial chain complex. This identification is tautological and you don't have to pass through singular homology to see it.
I hope that helps...
