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I am working on the following exercise: Show that a simplicial map induces a map on chain complexes (hence maps on homology between simplicial complexes).

Here is what I have so far...

Let $K, L$ be abstract simplicial complexes, $\phi: K \rightarrow L$ a simplicial map between them. Then for each $i \in \mathbb{N}$, $\phi$ induces a chain map $\phi_i : C_i (K) \rightarrow C_i (L)$ given by $$ \phi_i (\langle v_0, \ldots, v_i \rangle) = \left\{\begin{array}{ll}\langle \phi(v_0), \ldots, \phi(v_i) \rangle, & \mbox{ if the } \phi(v_j) \mbox{ are distinct} \\[30pt] 0, &\mbox{otherwise}.\end{array}\right. $$ To see that $\phi_i$ is a chain map, it suffices to check that $\phi_*$ commutes with the boundary maps $\partial_1 : C_i (K) \rightarrow C_{i-1} (K) $ and $\partial_2 : C_i (L) \rightarrow C_{i-1} (L) $, i.e., $$\partial_2 \circ \phi_i = \phi_i \circ \partial_1 : C_i (K) \rightarrow C_{i-1} (L).$$ Begin with the LHS first: $$ \partial_2 \circ \phi_i (\langle v_0, \ldots, v_i \rangle) = \left\{\begin{array}{ll} \sum_{j} (-1)^j \langle \phi(v_0), \ldots, \widehat{\phi(v_j)}, \ldots, \phi(v_i) \rangle, & \mbox{ if the } \phi(v_j) \mbox{ are distinct} \\[30pt]0, &\mbox{otherwise}.\end{array}\right. $$ I am STUCK on making sense of casework for the RHS, and I am posting this question to see if anyone visiting can help me in this endeavor. Any help would be greatly appreciated.

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Hint: The sign convention used in the definition of the boundary map will take care of the case $\phi(v_i) = \phi(v_j)$ for $i \neq j$ on the RHS of the equation. –  Alexander Thumm Feb 29 '12 at 10:51
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1 Answer 1

Here, according to our conventions the vertices on the left are in the proper order for K, but on the right we might have to introduce a sign in order to get the vertices in the proper order for L. If there is a repetition of vertices on the right, our conventions tell us the symbol stands for zero

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