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Consider we want to "calculate" the 1st simplicial homology group of torus without using any imagination or intuition related to chains and cycles and only by applying the algebra definitions involved such as the boundary operator and the quotient group.

How can this be done step by step?

In particular how to establish an isomorphism between the homology group obtaibed by the calculation and a group based on $\mathbb{Z}$ which in this case is the same as the fundamental group: $\pi_1(T^2)=\mathbb{Z} \oplus \mathbb{Z}$?

It seems to me that in most textbooks we have a jump in the proof or derivation by referencing to our visualization of what happens if we draw pictures of cycles on the surface of a torus. But I need to follow a proof that doesn't require us to imagine the boundaries and simplexes.

VVM
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    See for example here, or here. – Dietrich Burde Mar 18 '19 at 09:07
  • Most of the proofs will at some point give an argument such as "the attaching map is ... So this is the boundary operator". These arguments are mostly seen by drawing a triangulation usually. This is because there are many ways to define a torus, and many ways to parametrize it, but we want a uniform argument. However you could take your favourite definition of the torus (equation as a submanifold in $\mathbb{R}^3$, product of circles...) And work out the equations defining your triangulation in this context – Thibaut Benjamin Mar 18 '19 at 09:20

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From the duplicate: here are the boundary homomorphisms:

$$ \dots \rightarrow 0 \xrightarrow{d_3} \mathbb{Z} \xrightarrow{d_2} \mathbb{Z}^2 \xrightarrow{d_1} \mathbb{Z} \xrightarrow{d_0} 0$$

$d_0 = 0$

$d_1 = 0$ because the attaching map ($f = const.$) as there is one $0$-cell.

$d_2 = c_1 e_1^1 + c_2 e_2^1= 0$ because $f = ab a^{-1}b^{-1}$ so the coefficients are $c_i = +1 - 1$ respectively.

In particular it follows immediately that $$ H_1(T^2) = \ker \partial_1 / {\rm im }\, \partial_2 = \mathbb{Z} \oplus \mathbb{Z}. $$

Dietrich Burde
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  • Sorry but the attaching map and cell are not defined in the text that I am reading – VVM Mar 20 '19 at 06:41