If $X$ is a $CW-$complex. Are $C_*(X)$ and $C^{CW}_*(X)$ weakly equivalent? If $X$ is a $CW-$complex and we denote by $C_*^{CW}(X)$ the chain complexe given by $H_n(X_n,X_{n-1})$ in degree $n$ can we construct a weak equivalence $C^{CW}_*(X) \rightarrow C_*(X)$? I know their homologies are isomorphic but are they weakly equivalent? I'm pretty sure they are. 
A weak equivalence $\phi:C_* \rightarrow D_*$ is a chain map which induces isomorphism in homology in all degrees.
This is a harder problem to solve than if $X$ was a triangulated space viewed as a $CW-$complex since we have a canonical representative of the homology class that generates $H_n(\Delta^n, \partial \Delta^n)$, namely the identity map $\Delta^n \rightarrow \Delta^n$ but there is no such canonical choice for $H_n(D^n, \partial D^n)$. 
I don't know how to choose representatives of $H_n(D_e^n, \partial D_e^n)$ for all $n-$cells $D_e^n$ of $X$ for all $n$ in order to define a chain map $C_*^{CW}(X) \rightarrow C_*(X)$.
I appreciate any help with this question!
 A: First of all, as you point out, if you're dealing with triangulated spaces, the whole matter becomes easier. I suppose that's because you impose that the map $\partial \Delta^n \to X_{n-1}$ respects the triangulated structure, so everything works well when you look at the complexes in question. 
For "plain" CW-complexes, as we saw in the comments, I'm not sure there's a geometric proof. However, here's an algebraic proof that the two are weakly equivalent, in fact, they're homotopy equivalent. The proof is "stupid" in that it only relies on the fact that they're complexes of free abelian groups and have the same homology; and the map you get is not natural in $X$ in any reasonable sense (even with respect to cellular maps, while you could hope that it would be)
The proof is as follows: $C_*^{CW}(X)$ (resp. $C_*(X)$) are complexes of abelian groups, so they are weakly equivalent (in the sense of a zigzag of morphisms) to their homology (look for instance at the accepted answer here) ,therefore they are weakly equivalent to one another. 
This means they are isomorphic in the derived category $D_{\geq 0}(\mathbf{Ab})$. However, they are both chain complexes of free modules, so $\hom_{D_{\geq 0}(\mathbf{Ab})}(C_*^{CW}(X),C_*(X))$ is just the quotient of $\hom_{Ch_{\geq 0}(\mathbf{Ab})}(C^{CW}_*(X), C_*(X))$ by the homotopy relation, and similarly in the other direction. It follows that they are homotopy equivalent. 
Here's a possible geometric approach, to yield naturality : the category we will consider is a slight modification on CW-complexes : we will want to record how cells are attached, and morphisms will have to respect this (note that I'm not entirely sure that what I'm writing is correct, you should especially double check this bit - I'm writing it and correcting it at the same time as thinking about it.  Also, at the end, I don't get an actual conclusion, just a wild guess)
So an object in our category $C$ will be a CW-complex $X$ together with its "history" of construction, that is, for each $n$, a set $I_n$ and a family $\phi_i : S^n\to X^{(n)}$ of attaching maps. So essentially : a CW-complex, together with its CW-structure
A morphism between two such things will be in particular a cellular map, but actually the requirement will be stronger : a map $f: X\to Y$ will be a cellular map such that for all $n$, the map $X^{(n+1)}\to Y^{(n+1)}$ is induced by the map $X^{(n)}\to Y^{(n)}$ and a map $I_n\times D^{n+1}\to Y^{(n+1)}$ such that the composite with $Y^{(n+1)}\to Y^{(n+1)}/Y^{(n)}\cong \bigvee_{j\in J_n}S^{n+1}$ is, for each $i\in I_n$, just the quotient map $D^{n+1}\to S^{n+1} $ followed by the inclusion $S^{n+1}\to \bigvee_{j\in J_n}S^{n+1}$, for exactly one $j\in J_n$; and also such that $I_n\times D^{n+1}\to Y^{(n+1)}$ restricts to $I_n\times S^n\to J_n\times S^n$ with the induced map $I_n\to J_n$ and the identity $S^n\to S^n$
The goal is then to use the acylic models theorem. For notation, I will be following this statement . Our functor $F$ is $C_*^{CW}$, I think its definition is pretty clear (given that the maps are cellular but in fact send cells to cells, it's easy to see how it is defined on morphisms). Now I claim that $C_k^{W}$ is free on $\{D^k\}$, with the usual cell-decomposition : one $0$-cell, one $k-1$-cell to produce a $k-1$-sphere, and then one $k$-cell to fill it. 
Indeed, what is a map $D^k\to X$ in $C$ ? I claim that is the same data as a $k$-cell in $X$. Well clearly such a map determines a $k$-cell in $X$ : indeed look in degree $k$, you have, by definition of $C$, that $D^k\to X^{(k)}/X^{(k-1)}$ corresponds to picking precisely one $k$-cell (and since we required that this map is the quotient map, there is no additional data). Conversely, a $k$-cell of $X$ determines (obviously) a map $D^k\to X$. 
One may check that these two applications are inverse to one another (I think this uses the last requirement in my definition of $C$, that is, that a map $D^k\to X$ must respect the boundary : it clearly preserves the interior, because of the condition on quotient maps; and so to make sure we don't lose information one must impose that it preserves the boundary).
In any case $C_k^{CW}$ is free on $\{D^k\}$ (with the given cell-decomposition)
Then we put $V= C_*$, which is defined in the obvious way. We need to check that it is $k$ and $k+1$-acyclic at these models, which means that $H_k^{sing}(D^k),H_{k+1}^{sing}(D^k), H_k^{sing}(D^{k+1}), H_{k+1}^{sing}(D^{k+1})$ must be $0$ for $k>0$. Well this is just a classical fact about singular homology, and contractibility of $D^k$. 
It then follows that any natural transformation $H_0\circ C^{CW}_* \to H_0\circ C_*$ extends (uniquely up to homotopy) to a natural chain map $C_*^{CW}\to C_*$. It mustn't be hard to show that the isomorphism $H_0^{CW}(X)\to H_0(X)$ is natural, so we get our unique chain map that does this. 
My guess is that this chain map is a weak equivalence, but I'm not quite sure how to prove that. Note that this would provide some amount of naturality (although in a sense restricted : the maps of $C$ are quite restrictive)
A: At the request of Maxime Ramzi, I've copied my answer from a similar question:
Here is a nice zig-zag of quasi isomorphisms. Let $Sing(X)$ denote the realization of the singular set of X. Let $Song(X)$ denote the realization of the simplicial set of singular simplices that are cellular maps.
We have a chain of maps $X \leftarrow Song(X) \rightarrow Sing(X)$, where it is standard that these are weak equivalences and by design are cellular (where the latter spaces are CW complexes since they are realizations of simplicial sets). Hence, on CW chains these are quasi isomorphisms.
It is easy to see that CW and simplicial homology on the realization of a simplicial set coincide, so after taking CW chains we can extend to the right by an isomorphism of chain complexes getting us the simplicial chains of $Sing(X)$. Simplicial chains on $Sing(X)$ is exactly singular chains on X, so we are done.
