If X is a simplicial complex, are the complex of simplicial chains and the one of cellular chains of X identical? Let $X$ be a simplicial complex. The simplicial chain complex $C_*(X)$ is given by:
$$ C_k(X) := \{\mathbb{Z}-\textrm{combinations of oriented simplices in X, with the identification }\sigma = - \overline \sigma \}, $$
and the obvious boundary map ($\overline \sigma$ refers to the simplex $\sigma$ with the opposite orientation). The cellular chain is given by assigning to $X$ the obvious CW structure induced by the simplicial structure. We immediately get that we have a natural isomorphism:
$$ H_k(X_k, X_{k-1}) \sim C_k(X) $$
(where $X_i$ is the $i$--skeleton of $X$).
Hence we just need to know if the cellular boundary map: $H_k(X_k, X_{k-1}) \to H_{k-1}(X_{k-1}, X_{k-2})$ given by the "cellular boundary formula" (in hatcher's "algebraic topology" book) corresponds to the boundary map in simplicial homology.
Precise references would be much appreciated (indeed, it is rather curious that I wasn't able to find the answer somewhere for myself).
PS
note that this question refers to the "chain level". Obviously the two cellular and simplicial homologies are isomorphic!
PPS
Having thought a little bit about my question, I suspect that the answer is "obviously yes", and depends on the definition of cellular homology by itself, not on the cellular boundary formula!
 A: Yes, the cellular boundary formula agrees with the usual boundary map of simplicial homology, and this is immediate from the definitions.  Basically, the cellular boundary is just induced by the boundary map of singular chains, but the boundary of a singular simplex is defined as the alternating sum of faces in the exact same way as for simplicial homology.
Here are the details.  Consider a $k$-simplex $\sigma:\Delta^k\to X$ of $X$.  Then $\sigma$ (as a singular $k$-chain) represents an element of $H_k(X_k,X_{k-1})$ and we wish to show the cellular boundary of this element is the usual alternating sum of its faces, considered as singular $(k-1)$-chains representing elements of $H_{k-1}(X_{k-1},X_{k-2})$.  But this is immediate from the cellular boundary map: the cellular boundary map is the connecting homomorphism $H_k(X_k,X_{k-1})\to H_{k-1}(X_{k-1})$ composed with the canonical map $H_{k-1}(X_{k-1})\to H_{k-1}(X_{k-1},X_{k-2})$.  The connecting homomorphism by definition takes a cycle representing a relative homology class and takes its boundary.  So the connecting homomorphism takes the class of $\sigma$ in $H_k(X_k,X_{k-1})$ and sends it to the class of $\partial\sigma$, the alternating sum of its faces, in $H_{k-1}(X_{k-1})$.  The canonical map then just considers $\partial\sigma$ as a relative chain representing an element of $H_{k-1}(X_{k-1},X_{k-2})$.  Thus the cellular boundary of $\sigma$ is exactly the element of $H_{k-1}(X_{k-1},X_{k-2})$ given by the usual alternating sum of the faces of $\sigma$.
A: This is more subtle than it seems at first glance. Many years ago I asked myself why we consider the free abelian group $C'_k(X)$ generated by all oriented $k$-simplices of $X$ and then form the quotient
$$C_k(X) = C'_k(X)/\langle \sigma + \overline{\sigma} \rangle$$
in which $\sigma$ and $-\overline{\sigma}$ are identified.
The group $C_k(X)$ is free abelian with one generator for each $k$-simplex $s$ of $X$. In other words, why do we start with two generators for each $k$-simplex $s$ (the two oriented simplices corresponding to $s$) of $X$ if we actually only want one? 
The reason are the boundaries $\partial :  C'_k(X) \to C'_{k-1}(X)$. On the generators we define $\partial(\sigma) = \sum_{i=0}^k (-1)^i \sigma_i$, where the $\sigma_i$ are the faces of $\sigma$ with the induced orientations. Unfortunately there is no reasonable way to define $\partial(s)$ for an unoriented simplex $s$. The problem are the signs which we have to associate to the faces $s_i$ of $s$. There is no canonical way to do so. In fact, there are two valid options which stem from the two orientations of $s$. However, to bypass this problem we can make a  choice of an orientation for each $s$, subject to the condition that these choices are consistent over all simplices of $X$ (this means that if $s'$ is a face of a simplex $s$, then the orientation of $s'$ conincides with the orientation inherited from $s$). A nice way to do this is to choose a total ordering on the set of vertices of $X$. This gives us an ordering of the vertices of each simplex $s$, i.e. we can write $s = (p_0,\dots,p_k)$ in a unique order. This allows to define $\partial(s) = \sum_{i=0}^k (-1)^i s_i$, where $s_i$ is obbtained from $s$ be by deleting $p_i$. Let us denote the resulting chain complex by $C^{ord}_k(X)$. Recall that it involves a choice to form it. 
It is a nice excercise to show that there exists a natural isomorphism $C^{ord}_*(X) \to C_*(X)$.
There is a similar issue with the cellular chain complex $D_k(X) = H_k(X_k,X_{k-1})$. These groups are free abelian with one generator for each $k$-simplex of $X$, but again it requires suitable choices to identify them with the generators of $C_k(X)$. To understand this let us look into the chapter about cellular homology in Hatcher's book "Algebraic topology", http://pi.math.cornell.edu/~hatcher/AT/AT.pdf. The boundaries can be computed via the "cellular boundary formula". This formula involves the characteristic maps of the $k$-cells $e_s$ in the form $\phi_s : S^{k-1} \to X_{k-1}$ (here $s$ is any $k$-simplex of $X$). But we only have a canonical attaching map $\iota_s : \partial \lvert s \rvert \hookrightarrow X_{k-1}$, where $\lvert s \rvert \subset\lvert X \rvert$ is the geometric realization of $s$. To get the characteristic maps in the desired form, we need a specific identification of $\lvert s \rvert$ with $D^k$.  We may assume that $D^k$ is the standard $k$-simplex in $\mathbb{R}^{k+1}$ whose vertices are the standard base vectors $e_i$, but there are many homeomorphisms $D^k \to \lvert s \rvert$, so this involves a choice. To do this, we choose again a total ordering on the set of vertices of $X$. Then for each $s = (p_0,\dots,p_k)$ we get a canonical affine homeomorphism $h_s : D^k \to \lvert s \rvert$ such that $h_s(e_i) = p_i$. Then define $\phi_s = \iota_s h_s \mid_{S^{k-1}}$. For the sake of clarity let us write $D^{ord}_*(X)$ for the resulting cellular chain complex.
Now $X_k/X_{k-1}$ is a wedge of $k$-spheres and thus each $\phi_s$ determines a generator of $D^{ord}_k(X)$ by identifying $S^k$ with one of spheres in this wedge ($\phi_s$ induces $\phi'_s : S^k = D^k/S^{-1} \to X_k/X_{k-1}$ and we take $(\phi'_s)_*(g_k) \in D^{ord}_k(X)$, where $g_k$ is a generator of $H_k(S^k)$).
Now define isomorphisms $f_k : C^{ord}_k(X) \to D^{ord}_k(X), f_k(s) = (\phi'_s)_*(g_k)$. It is a nice exercise (based on Hatcher) to verify that the $f_k$ constitute a chain map $f_* : C^{ord}_*(X) \to D^{ord}_*(X)$.
