Let $K$ be a simplicial complex: it consists from the set (called the set of vertices) and a family of subsets of set of vertices satisfying the property of being closed under taking subsets (those set are called simplices: condition translates that the subset of a simplex is again simplex). Homology of simplicial complex is defined as follows: we consider $C_q(K)$, the free $R$-module ($R$ is a ring) generated by all oriented $q$ simplices with relation $\sigma+\sigma'=0$ where $\sigma'$ is $\sigma$ with reversed orientation. The boundary is defined with as $\partial=\sum_{i=1}^q (-1)^i\partial_i$ where $\partial_i[v_0,...,v_q]=[v_0,...,v_{i-1},v_{i+1},...,v_q]$.

Now given any abstract simplicial set $X=(X_q)_{q=0}^{\infty}$ one can consider $S_q(X)$=free module generated by $X_q$. The boundary is defined as $\partial=\sum_{i=0}^q(-1)^i\partial_q$ where $\partial_i$ are part of simplicial set data.

If we start with simplicial complex one can associate simplicial set $SimpSet(K)=(Ss_q(K))_{q=0}^{\infty}$ where $Ss_q(K)=\{(v_0,...,v_q): [v_0,...,v_q] \in K \}$ where $(v_0,...,v_q)$ is an ordered tuple and we allow repetitions. Face and degeneracies are defined as follows: $$\partial_i (v_0,...,v_q)=(v_0,...,v_{i-1},v_{i+1},...,v_q)$$ $$\sigma_i (v_0,...,v_q)=(v_0,...,v_i,v_i,v_{i+1},...,v_q).$$ This defines simplicial set.

Therefore we arrive at two a priori different homology theories but at the end of the day they should coincide. My guess for a natural (chain) map yielding isomorphism in cohomology would be $[v_0,...,v_q] \mapsto (v_0,...,v_q)$.

How to prove that this map induces isomorphism in homology? If it is not the case, how this isomorphism should be defined?

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    $\begingroup$ The chain complex of the $C_q(K)$ should agree up to isomorphism with the normalized Moore complex of the associated simplicial R-module under the Dold-Kan correspondence, I believe, but I'm not certain enough to put it as an answer. If this is the case, it's actually a point-set isomorphism rather than just a quasi-isomorphism. $\endgroup$ Nov 30, 2017 at 0:21

1 Answer 1


There is a problem with your definition of the boundary map $\partial : C_q(K)\to C_{q-1}(K)$. The formula $\partial _i[v_0,\cdots,v_q]=[v_0,\cdots,v_{i-1},v_{i+1},\cdots,v_q]$ depends on an ordering of the vertices of the simplex $[v_0,\cdots,v_q]$, but such an ordering is not part of the definition of a simplicial complex. One can try to avoid this problem by including orientation data in the definition of $C_q(K)$ as you have done, but then one has to specify how an orientation of a $q$-simplex induces orientations of each of its $(q-1)$-dimensional faces, in order for the alternating sum formula for the boundary map to make sense. This can be done, but it takes a little work. (For example, one would have to verify that $\partial^2=0$, which takes more work if one only has orientations and not orderings of vertices.)

Your proposed chain map $[v_0,\cdots,v_q]\mapsto (v_0,\cdots,v_q)$ is also not well defined for the same reason. If one chooses a partial ordering of all the vertices of $K$ that restricts to a linear ordering on the vertices of each simplex, then one gets a well-defined chain map, but it is not natural since it depends on the choice of the ordering. To get a natural map one needs to replace simplicial complexes with ordered simplicial complexes or more generally semi-simplicial sets, which are simplicial sets without degeneracies.

After making these adjustments to the definitions one does obtain a well-defined chain map. This can be shown to induce an isomorphism on homology by first treating the case of finite complexes by induction using the five lemma applied to Mayer-Vietoris sequences. Then one passes to infinite complexes using a direct limit argument.


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