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 $[v\_0,...,v\_q] \mapsto (v\_0,...,v\_q)$. 



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> 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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Homology of simplicial complex versus homology of simplicial _set_
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. 


