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Let $R = {\bf Z}$

Let $\partial_i : C_i \rightarrow C_{i-1}$ be a boundary map

where $C_{-1} = \{ 0 \}$, $C_i$ is the set of all maps $f$ and $f: \Delta_i \rightarrow M$. Let $Z_n = $Ker of $\partial_n$

If $f $ is a loop in $M$ then $\partial_1(f)=0$.

If $M$ is $S^2$, and if $f$ is in $Z_2$, what is $f$ ?

I want to know the example for $f$.

In fact, if $f$ maps the boundary of $\Delta_2$ into a point in $M$, i.e., $\Delta_2$ wraps $M $ one time, $f$ is in $Z_2$. However does it imply that $\partial_2 (f) =0$ ?

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up vote 4 down vote accepted

If $f$ is as you describe, and $1,2,3$ denote the three endpoints of $\Delta_2$ then $\partial_2 f = f_{23} - f_{13} + f_{12} = \ast - \ast + \ast = \ast$ where $\ast \in M$ is the point you map the boundary to and $f_{ij}$ is $f$ restricted to the boundary path from $i$ to $j$. Hence the answer is no, in this case $f$ is not an element of $\mathrm{ker} \partial_2$.

Note that your notation is not standard. Commonly, $C_i$ is used to denote the $i$-th chain group which is the set of all formal sums of $i$-simplexes. Hence if $\sigma_i : \Delta_i \to M$ is a continuous map then an element of $C_i$ looks like $\sum_{k=1}^n c_k \sigma_k $ for $c_k \in \mathbb Z$.

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