Given a triangulable space $X$, by definition there exists an homeomorphism $F:|K|\to X$, where $|K|$ is the support of a simplicial complex. It's easy to see that $X$ inherits a natural structure of CW-complex. Now, I have some problems with the proof my professor gave us of the fact that for all $q\in\mathbb Z$: $$H_q^\Delta(K)\cong H_q^{CW}(X)$$
The proof is the following.
Let fix an orientation on the vertices of $K$, such that every simplex of $K$ inherits an orientation. It is clear that for every $\tau=\langle P_0,\ldots,P_q\rangle\in K^q$ (here $K^q$ is the $q$-skeleton of $K$) there exists an affine map $\hat\tau:\Delta_q:=\langle e_0,\ldots,e_q\rangle\to K$ such that $\hat\tau(e_i)=P_i$ for every $i=0,\ldots,q$. This allows us to define: $$\Phi^q_\tau:\mathbb D^q\xrightarrow{\psi_q}\Delta_q\xrightarrow{\hat\tau}K\xrightarrow{F}X$$
where $\mathbb D^q$ is the closed $q$-dimensional disk and $\psi_q$ is an homeomorphism between $\mathbb D^q$ and the standard $q$-simplex $\Delta_q$. Thus for every $q$ we can consider $$X^q:=\bigcup_{q'\leq q\\\tau\in K^{q'}}\Phi^{q'}_\tau(\mathbb D^{q'})$$ which can be proved to be the $q$-skeleton of the CW-complex structure induced on $X$. At this point any $\tau\in K^q$ induces the map $F_*\tau:=F\circ\hat\tau:\Delta_q\to X^q$, which is an element of $S_q(X^q)$, where $S_q(X^q)$ is the free module generated by the singular $q$-simplices. So for every $q\in\mathbb Z$ it's possible to define $u_q:C_q(K)\to S_q^{CW}(X):=H_q(X^q,X^{q-1})$ in the following way:
$$u_q:\langle\tau\rangle\in C_q(K)\to F_*\tau\in S_q(X^q)\xrightarrow{j_q}j_q(F_*\tau)\in S_q(X^q,X^{q-1})\to [j_q(F_*\tau)]\in H_q(X^q,X^{q-1})=:S_q^{CW}(X)$$
It's briefly said that all the $u_q$ are clearly isomorphisms of modules, but I don't see why (and this is my first problem with the proof). Now, given that for true, if we prove that the following diagram commutes $\require{AMScd}$
\begin{CD}
C_q(K) @>{u_q}>> S_q^{CW}(X)\\
@VVV @VVV\\
C_{q-1}(K) @>{u_{q-1}}>> S_{q-1}^{CW}(X)
\end{CD} we are done, beacuse we have an isomorphism at a chain level, which, thanks to the functoriality of homology, induces an isomorphism in homology. Now:
$$\begin{align*}&u_{q-1}(\partial \tau)=[j_{q-1}(F_*\hat{\partial\tau})]=[j_{q-1}\partial(F_*\hat\tau)]\\&d_q(u_q(\tau))=d_q[j_q(F_*\hat\tau)]=j_{q-1}(\partial_*)_q[j_q(F_*\hat\tau)]\end{align*}$$ where the $j_{q-1}$ outside the squared parenthesis is the morphism that goes from $H_{q-1}(X^{q-1})$ to $H_{q-1}(X^{q-1},X^{q-2})$ and $(\partial_*)_q:H_q(X^q,X^{q-1})\to H_{q-1}(X^{q-1})$.
I would be very glad if someone could explain me the reasons why the $u_q$'s are isomorphisms and the diagram written above commute, because I can't see them. I appreciate your help and I thank you in advance.
Edit:
I'm adding some further details. As specified in the comments, given a CW-complex $X$, we used the fact that $\frac{X^j}{X^{j-1}}\cong \bigvee_{\alpha\in A_j}S^j_\alpha$ in order to define the modules of the cellular chain complex as: $$S_q^{CW}(X):=H_q^{\text{sing}}(X^q,X^{q-1})\cong\bigoplus_{\alpha\in A_q} \tilde H_q(S^q_\alpha)=\bigoplus_{\alpha\in A_q}R\cdot[e^q_\alpha]$$ where $R$ is the ring of the coefficients and $e^q_\alpha$ a $q$-cell. At this point the differentials $d_q:S_q^{CW}(X)\to S_{q-1}^{CW}(X)$ are the ones induced by the following diagram:
Best Answer
As you write in your question the $q$-cells of $X$ correspond bijectively to the $q$-simplices of $K$ and for $\tau$ a $q$-simplex the $q$-cell $e_\tau^q$ is has characteristic map $\Phi^q_\tau$. Consider the composite $$C_q(K)\xrightarrow{u_q} H_q(X^q, X^{q-1})\xrightarrow{\cong} H_q(X^q/X^{q-1},\ast) = \tilde{H}_q(X^q/X^{q-1}).$$ A careful check shows that $\tau\in C_q(K)$ is mapped to the singular simplex $$\Delta^q\xrightarrow{\hat\tau} K^q\xrightarrow{F} X^q\to X^q/X^{q-1},$$ which is exactly the basis element of $\tilde H_q(X^q/X^{q-1})$ corresponding to $e_\tau^q$. Thus, $u_q$ maps basis elements bijectively, hence is an isomorphism.