Every time I look up something about singular homology I seem to find a different definition, so I just want to clear up a couple of things. Let $X$ be a topological space and $\Delta^n$ the $n$-th standard simplex. An $n$-simplex is then a continuous map $\sigma \colon \Delta^n \to X$. For the sake of simplicity (no pun intended), let us set $X=\mathbb{C}\setminus\{0\}$ and $\sigma \colon [0,1] \cong \Delta^1 \to X, t \mapsto e^{2\pi i t}$.

$(1):$ Some authors define the group of $n$-cycles $C_n(X,\mathbb{Z})$ as the free abelian group generated by all the $n$-simplices. Others (for example Wikipedia) as the free abelian group generated by the images of all $n$-simplices. Now, this second definition does not seem fitting to me: in our example, the double loop $\sigma^2$ around zero has the same image as the single loop $\sigma$, and we clearly do not want to identify the two maps, do we?

$(2):$ In any case, we are talking of a free abelian group of $n$-cycles. So all operations are formal, the only concrete things we have are the generators, i.e., the $n$-simplices. Hence, in the above example, a double loop around the origin is not, strictly speaking, the same as twice a single loop around the origin: $\sigma^2 \neq 2\sigma$. But now it seems to me that we do want to identify these, don't we?

$(3):$ Related to $(2)$. If, on the other hand, we identify cycles having the same image and boundary (hence $C_n(X,\mathbb{Z})$ is a non-trivial quotient of the free abelian group), then each $0-$cycle is $0$, because it is identified with its additive inverse. We do not want this, do we?

So, here it all is. I am very grateful for any comment about correct or erroneous reasonings and for any good references. Thanks in advance!

  1. No author should define it as the free abelian group generated by the images, and wikipedia doesn't do this.

EDIT: I've erased a big part of this point of the answer, since I was supposing that wikipedia didn't define it using the images. However, as said in the comments, wikipedia explicitly states that

The basis for the group is the infinite set of all possible images of standard simplices.

Thus, my interpretation of what was in wikipedia is incorrect: it indeed defines as the free abelian group generated by the images, which is incorrect.

  1. Correct, it is not the same. However, when you go to homology they will be (it may be a good exercise to prove this. An even better one to prove this only with a drawing).
  2. Although $\sigma+\sigma=``\sigma^2"$, cycles with same image will not necessarily be equal in homology. For instance, $``\sigma^2"$ and $\sigma$, where $\sigma$ is $t \mapsto e^{2\pi it}$ in $S^1$, have the same image, but are different in homology.
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  • $\begingroup$ why would they be different in homology? The images of each in the kernel will be the subtraction of their boundary points, and they represent the same coset in homology since they don't bound a higher dimensional simplex. I.e: aren't they homologous? $\endgroup$ – Andres Mejia Feb 25 '18 at 18:55
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    $\begingroup$ @AndresMejia $[\sigma]$ is a generator for $H_1(S^1)$. $2 [\sigma] \neq [\sigma]$. $\endgroup$ – Aloizio Macedo Feb 25 '18 at 18:59

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