Real singular chain complex

For every topological space $X$, use Continuous singular simplex and real singular chain group and the real-linear instance of The singular boundary operator: ${\mathrm{\partial }}_k[\sigma ]={\sum }_{i=0}^k(-1{)}^i[\sigma \circ {\delta }_i](k\ge 1),{\mathrm{\partial }}_k=0(k\le 0)\mathrm{.}$ Each sum is finite. The real singular chain complex is $C_{\bullet }(X;\mathbb{R})$ with these differentials. In degree one, a path has boundary $[\sigma (1)]-[\sigma (0)]$.

For completeness, if $k\ge 2$, in the expansion of ${\mathrm{\partial }}_{k-1}{\mathrm{\partial }}_k[\sigma ]$, a pair of omitted vertices $i<j$ occurs as $\sigma {\delta }_j{\delta }_i$ and $\sigma {\delta }_i{\delta }_{j-1}$. The affine maps insert zeros at the same two positions and leave all other coordinates in order, so they are equal. Their coefficients $(-1{)}^{i+j}$ and $(-1{)}^{i+j-1}$ cancel. Every term belongs to exactly one such pair. Thus ${\mathrm{\partial }}^2=0$ on every generator and hence on every finite chain. For $k=1$ the composite is zero because ${\mathrm{\partial }}_0=0$; all lower groups or maps are zero. This verifies well-definedness as an unaugmented chain complex directly.

Repeated or constant faces are counted with their signed multiplicities; no nondegeneracy assumption is imposed. For a point the boundary coefficient is ${\sum }_{i=0}^k(-1{)}^i$, equal to $1$ for positive even $k$ and $0$ for odd $k$, while ${\mathrm{\partial }}_0=0$. For the empty space the entire complex is zero. No choice is used.