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I am reading James Munkres' book, Elements of Algebraic Topology.
Theorem 6.2 on page 35 concerns the homology groups of the 2dimensional torus.
Munkres shows that [itex] H_1 (T) \simeq \mathbb{Z} \oplus \mathbb{Z} [/itex] and [itex] H_2 (T) \simeq \mathbb{Z} [/itex].
After some work I now (just!) follow the proof that [itex] H_1 (T) \simeq \mathbb{Z} \oplus \mathbb{Z} [/itex] but I need some help to understand a point in the proof of [itex] H_2 (T) \simeq \mathbb{Z} [/itex].
Munkres' argument to show [itex] H_2 (T) \simeq \mathbb{Z} [/itex] is as follows:

To compute [itex] H_2 (T) [/itex], note that by (2) any 2cycle d of T must be of the form [itex] p \gamma [/itex] for some p. Each such 2chain is in fact a cycle,by (4) , and there are no 3chains for it to bound. We conclude that
[itex] H_2 (T) \simeq \mathbb{Z} [/itex]
and this group has as generator the 2cycle [itex] \gamma [/itex].

I do not fully understand why any 2cycle d of T must be of the form [TEX] p \gamma [/TEX] for some p.
Can someone please explain exactly why this follows?
Would appreciate some help.

To give members of the Physics Forums the context of the post above, the text of Theorem 6.2 and its proof follow:
Some of my thoughts ... ...
Basically, to show that any 2cyclce of L (i.e. T) os of the form \(\displaystyle p \gamma \), we have to show the following:
If [itex] d = \sum_i n_i \sigma_i [/itex] where [itex] \partial d = 0 [/itex] then [itex] d = p \gamma [/itex].
We have, of course that [itex] \gamma = \sum_i \sigma_i [/itex]
Note that we have that if d is a 2chain of L and if [itex] \partial d [/itex] is carried by A then d is a multiple of [itex] \gamma [/itex].
Munkres defines 'carried by' in the following text taken from page 31:
Hope someone can help.
Peter
Theorem 6.2 on page 35 concerns the homology groups of the 2dimensional torus.
Munkres shows that [itex] H_1 (T) \simeq \mathbb{Z} \oplus \mathbb{Z} [/itex] and [itex] H_2 (T) \simeq \mathbb{Z} [/itex].
After some work I now (just!) follow the proof that [itex] H_1 (T) \simeq \mathbb{Z} \oplus \mathbb{Z} [/itex] but I need some help to understand a point in the proof of [itex] H_2 (T) \simeq \mathbb{Z} [/itex].
Munkres' argument to show [itex] H_2 (T) \simeq \mathbb{Z} [/itex] is as follows:

To compute [itex] H_2 (T) [/itex], note that by (2) any 2cycle d of T must be of the form [itex] p \gamma [/itex] for some p. Each such 2chain is in fact a cycle,by (4) , and there are no 3chains for it to bound. We conclude that
[itex] H_2 (T) \simeq \mathbb{Z} [/itex]
and this group has as generator the 2cycle [itex] \gamma [/itex].

I do not fully understand why any 2cycle d of T must be of the form [TEX] p \gamma [/TEX] for some p.
Can someone please explain exactly why this follows?
Would appreciate some help.

To give members of the Physics Forums the context of the post above, the text of Theorem 6.2 and its proof follow:
Some of my thoughts ... ...
Basically, to show that any 2cyclce of L (i.e. T) os of the form \(\displaystyle p \gamma \), we have to show the following:
If [itex] d = \sum_i n_i \sigma_i [/itex] where [itex] \partial d = 0 [/itex] then [itex] d = p \gamma [/itex].
We have, of course that [itex] \gamma = \sum_i \sigma_i [/itex]
Note that we have that if d is a 2chain of L and if [itex] \partial d [/itex] is carried by A then d is a multiple of [itex] \gamma [/itex].
Munkres defines 'carried by' in the following text taken from page 31:
Hope someone can help.
Peter
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