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?