Question about Hurewicz theorem There's a step of Hurewicz theorem (cited here) that I'd like to clear. I'd to know whether my thoughts are correct.
Briefly to introduce the notation, the steps are:

*

*We can suppose without loss of generality that $X$ is a $CW$ complex as a consequence of $CW-$approximation theorem since weak homotopic equivalence induces isomorphism in homology. We can say something more: we can replace $X$ by a $CW$ such that $X^{n-1} = \left\lbrace*\right\rbrace$ (here $X^i$ will denote the skeletons of the complex).


*$\pi_{n}(X^{n+1})\simeq \pi_{n}(X)$(by $X$ $(n-1)$-connected) and $H_{n}(X^{n+1})\simeq H_{n}(X)$ due to standard facts about homology of $CW$ complex. So it's enought to prove the theorem for $X^{n+1}$.


*We can easily see that $X^{n} = \bigvee\limits_{k} \mathbb{S}_{k}^n$.
Here it comes the question:
Now, called $\varphi_j$ the attaching map of the $(n+1)$-cells, we can consider a map $\varphi : \bigvee\limits_{j} \mathbb{S}_{j}^n \longmapsto X^n = \bigvee\limits_{k} \mathbb{S}_{k}^n$.
First question: who is this map? I think it should be the union of $\varphi_j$ so it should start from $\bigsqcup\limits_{k} \mathbb{S}_{k}^n$ as a standard attaching map, so why it doesn't?
Second question: Now I'd like to prove that the mapping cone of $\varphi$ i.e $C_\varphi = (C  \bigvee\limits_j \mathbb{S}_{j}^n\sqcup X^n)/(x,1)\sim \varphi(x)$ is exactly $X = X^{n+1}$.
In other words since $X^{n+1}$ is constructed a pushout, that $C_\varphi = (X^{n} \sqcup \bigsqcup_j \mathbb{D}_j^{n+1})/x \sim \varphi(x)$.
What concerns me is that even if the cone of the space commute with the wedge sum, we have that $C_{\varphi} = (\bigvee\limits_j \mathbb{D}_{j}^{n+1}\sqcup X^n)/(x,1)\sim \varphi(x)$ which doesn't seem the definition of $X^{n+1}$ I'm looking for since here I'm dealing with $\bigvee\limits_j \mathbb{D}_{j}^{n+1}$ and not with $\bigsqcup\limits_j \mathbb{D}_{j}^{n+1}$ I was hoping. What's wrong with the attaching maps here? What I'm missing?
Thoughts: I think I have an explanation of this fact, but I'd like to give myself a neat proof as possible. I thought that since $\varphi_j(\mathbb{D}_j^{n+1})$ is a subcomplex of $X$(it's a closed subset of $X$ which is a union of closed cells), we must have, since $X^{n+1}$ is built by attaching some $(n+1)-$cells to a very special space $X^n$ in this case, i.e $\bigvee\limits_{k} \mathbb{S}_{k}^n$, that the $0$ skeleton of the subcomplex $\varphi_j(\mathbb{D}_j^{n+1})$ coincide with the $0$-skeleton of the complex, i.e the point of the bouquet of spheres. So in reality the attaching map pass to the quotient inducing a map from $\bigvee\limits_j \mathbb{D}_{j}^{n+1}$, (possibly as an homeomorphism between $C_\varphi$ and $\bigvee\limits_j \mathbb{D}_{j}^{n+1}\sqcup X^n)/(x,1)\sim \varphi(x)$.
In my reasoning correct? Any help, solution or reference in order to prove the two questions are appreciated, as well as any answer.
 A: You are more or less right.
$\varphi_i: S^n \to X$ maps fixed point ($0$-skeleton) of $S^n$ to the only $0$-cell of $X^n$. Together they define a map $\coprod S^n \to X$, but since all fixed points of all spheres have the same image, this maps factors through the wedge: $\coprod S^n \to \bigvee S^n \to X$.
For discs it's better to look at the picture. For one $\varphi_i,$ "mapping cone of $\varphi_i$" is a fancy way to call the result of attaching a disc along $\varphi_i$ (though not as fancy as "homotopy colimit"). Attaching a disc along each $\varphi_i$ was giving us $X^{n+1}$. This is how it looks if we take two distinct mapping cones:

Now suppose you first take the map $\bigvee S^n \to X$ and form its cone. It will now look like that:

(all the tips of cones are now glued together in one point, which I denote by $\bar{x_0})$ The segment $x_0\bar{x_0}$ can be contracted onto $x_0$. What we get then is the previous picture, with discs attached along each $\varphi_i.$
So you can safely assume that you glue a wedge of discs along a wedge of circles when building $X^{n+1}$ from $X^n$.
