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In my previous question, I asked how one could say that the elements of the $p$-th singular chain group with coefficients in an abelian group, $\Delta_p(X;G) := \Delta_p \otimes_{\mathbb Z} G$, $G \in \mathsf {Ab}$, can be written as formal sums $$\sum_\sigma a_\sigma \sigma,$$ where $a_\sigma \in G$. In the comments, with the help of peter a g, I arrived at the conclusion that $$\Delta_p(X;G) \simeq \bigoplus_{i\in I}G,$$ where $I$ is the set of all singular $p$-simplices on $X \in \mathsf {Top}$. My question is now, how do we identify the direct sum above with the set of formal $G$-linear combinations of $p$-simplices?

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  • $\begingroup$ What does an element in the direct sum look like? There’s a copy of $G$ per singular simplex. So for each simplex, look at what element of $G$ is “chosen” and then just formally add up these for all the simplifies and you’ve got yourself a chain. One nuance - this is the direct sum so there are only finitely many non-0 terms, so only finite sums work which is good for intuition! $\endgroup$ Jun 18, 2021 at 16:59
  • $\begingroup$ Okay, that makes a little bit of sense $\endgroup$ Jun 18, 2021 at 17:10
  • $\begingroup$ So for example, assume that you have a countable set of singular simplices (this is unlikely to be true for most spaces but will help illustrate), take some ordering, then an element of the direct sum is something like $(g_1, g_2, 0, 0, g_5...)$. The singular chain this corresponds to is $g_1 \sigma_1 + g_2 \sigma_2 + g_5 \sigma_5...$ $\endgroup$ Jun 18, 2021 at 18:54
  • $\begingroup$ Yes, we can multiply the elements of $G$ by singular simplices, because they are represented by functions from the set $I$ of simplices to $\mathbb Z$, correct? $\endgroup$ Jun 18, 2021 at 19:16
  • $\begingroup$ It's more of just two ways of writing elements down once you have a basis. Think of linear algebra and the vector $(1,0,1) = 1e_1 + 0e_2 + 1e_3$. Does this make sense? If you need more explanation, I can write an answer $\endgroup$ Jun 18, 2021 at 20:57

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You've said that you've arrived at the conclusion that $\Delta_p(X;G) \cong \bigoplus_{i \in I} G$. That is, to each singular simplex $\sigma$, we assign a copy of $G$. An element of this direct sum is given by choosing an element of $G$ for each $i$, such that only finitely many of the elements chosen are non-zero. In other words, each element gives us a function $I \to G$, where only finitely many of the image of $I$ are non-zero. We can call this function $g:I \to G$ and denote $g(i) = g_i$.

Now, for each element in $\Delta_p$, you can write the corresponding formal sum by $\sum g(i) i = \sum g_i i$. The formal sum really is just associating each simplex with its weight in $G$.

As I mentioned in my comments above, this is no different than looking at $(a,b,c) \mapsto ae_1 + be_2 + ce_3$.

I think you may be looking too hard into the "multiplication" $g \cdot \sigma$. It really is just a weight / a "formal" multiplication. A free $G$-module by definition is just the $G$-module generated by a set, so the module multiplication is just putting them next to each other.

In the $\mathbb{Z}$ case, $3 \sigma$ doesn't mean much either. You can think of it as 3 copies of the simplex, but in the same way you can think of $g$ copies fo the simplex in a formal sense.

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  • $\begingroup$ Thanks for your help! $\endgroup$ Jun 18, 2021 at 22:31

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