Is the homology of the chain complex of singular simplices in a topological group with point wise addition the usual homology? An, at first sight, surprising exercise in the higher homotopy groups is that if your space is a topological group then addition in the higher homotopy groups is the same as point-wise multiplication of representative maps (where we are thinking of elements of $\pi_n(X)$ as maps $(I^n, \partial I^n) \to (X,x_0)$ where $x_0$ is the identity element of $X$, and addition is done in the last component of $I^n$).
Now I'm wondering: is there any analogous result for singular homology?
If $X$ is a topological group then we have a function $ev: C_n(X) \to C_n(X)$ given by $ev(\sum_{i=1}^m \alpha_i \sigma_i)(t) =\sum_{i=1}^n \alpha_i \sigma_i(t)$ for $t \in \Delta^n$. Here $\sigma_i:\Delta^n \to X$ and $\alpha_i = \pm 1$ so $\sum_{i=1}^m \alpha_i \sigma_i$ is a generic element of $C_n(X)$. The only ambiguity in how we chose to express an element of $C_n(X)$ in this way is the order of the sum and canceling pairs, and this considered, $ev$ is still well defined.
The best I can hope for here is that $ev$ sends homologous cycles to homologous cycles. This seems to work for the circle at least with reduced homology. 
I believe an equivalent result would be that in a topological group we can redefine the chain complex so that $C_n(X)=\{\sigma: \Delta^n \to X\}$, with addition of simplices being point-wise addition of maps and boundary maps now $ev\circ \partial$, and the homology would still be the same. Note that the zero element of $C_n(X)$ would be the constant map at the identity element. This is a legitimate chain complex because $ev \circ \partial \circ ev \circ \partial=ev \circ \partial \circ \partial$
using the relation $ev \circ \partial \circ ev =ev \circ \partial$, although a more convincing argument may be to just run through the argument that $\partial^2 =0$ applying $ev$ at every step and observing that the same cancellations happen.
If there isn't any analogous result then I would like to know why not.
 A: You can do what you suggest: if $X$ is a topological abelian group, then you can form a chain complex $A_\bullet$ where $A_n$ is the set of all maps $\Delta^n\to X$, which form a group under pointwise addition of maps.  The boundary map is given by taking the usual boundary of an $n$-simplex as an alternating sum of $(n-1)$-simplices and then adding those $(n-1)$-simplices together pointwise, just as you suggest with your $ev$ map.
However, the homology of this chain complex $A_\bullet$ is not the homology of the space $X$!  Instead, $H_n(A_\bullet)$ is naturally isomorphic to the $n$th homotopy group $\pi_n(X)$.  The proof is a bit complicated but essentially amounts to showing that in computing the homology of $A_\bullet$, you can restrict to just those simplices which map all but one of their boundary faces to the identity element of $X$.  When you restrict to such simplices, a cycle is then just a simplex $\Delta^n\to X$ which maps the entire boundary $\partial\Delta^n$ to the identity element of $X$, which induces a map $\Delta^n/\partial\Delta^n\cong S^n\to X$, and thus an element of $\pi_n(X)$.  You can find the details in Section III.2 of Goerss and Jardine's Simplicial Homotopy Theory (the specific result is Corollary 2.7, taking $A$ to be the singular simplicial set of $X$).
