Homotopic maps and attaching spaces Let $f, g : \mathbb{S}^{n-1} \to X$ be two continuous maps from the sphere into a compact and Hausdorff space $X$. I want to show that if $f$ and $g$ are homotopic, then attaching an $n$-cell to $X$ via $f$ or via $g$ yields spaces with the same homotopy type: $D^n \cup_f X \cong D^n \cup_g X$.
This has been answered in this post, but I didn't completely understand the answer. So I would like some help in the following steps:
I know that there is a deformation retraction $r : (D^n \times I) \to (D^n \times \{0\}) \cup (\mathbb{S}^{n-1} \times I)$, where $I = [0,1]$.
Let $H : \mathbb{S}^{n-1} \times I \to X$ be a homotopy between $f$ and $g$ and let $$\pi : (D^n \times I) \,\dot{\cup} \, X \to (D^n \times I) \cup_H X$$ and $$\rho : ((D^n \times \{0\}) \cup (\mathbb{S}^{n-1} \times I)) \, \dot{\cup} \, X \to ((D^n \times \{0\}) \cup (\mathbb{S}^{n-1} \times I)) \cup_H X $$
be the projection maps. 

  
*
  
*How do I show that $r$ descends to the quotient to a deformation retraction $$R : (D^n \times I) \cup_H X \to ((D^n \times \{0\}) \cup (\mathbb{S}^{n-1} \times I)) \cup_H X \,\,?$$
  
*How do I show that $$((D^n \times \{0\}) \cup (\mathbb{S}^{n-1} \times I)) \cup_H X$$ is homeomorphic to $$D^n \cup_f X \,\, ?$$

PS.: I know nothing about cofibrations.
 A: *

*Let $B \subset Y \subset Z$ with inclusion $i : Y \to Z$ and let $r : Z \to Y$ be a strong deformation retraction. Let $f : B \to X$ be a map. Then we get the two adjunction spaces $Y \cup_f X$ and $Z \cup_f X$ which are quotients of $Z \dot{\cup} X$ and $Y \dot{\cup} X$ with respct to the equivalence relation $\sim$ generated by $b \sim f(b)$ for $b \in B$. The map $r \dot{\cup} id : Z \dot{\cup} X \to  Y \dot{\cup} X$ respects $\sim$ and thus induces a map $r' : Z \cup_f X \to Y \cup_f X$ which is obviuously a retraction. A similar argument shows that a deformation $D : Z \times I \to Z$, $D : id \simeq ir$ rel. $Y$ induces a deformation $D' :  (Z \cup_f X) \times I \to  Z \cup_f X$, $D' : id \simeq i'r'$. This is true because if $p$ is quotient map, then so is $p \times id_I$.


*Define $\iota : D^n \to D^n \times \{ 0\}\ \cup S^{n-1} \times I, \iota(x) = (x,0)$. The map $\iota \dot{\cup} id$  respects $\sim$ and thus induces a map $\iota' : D^n \cup_f X \to (D^n \times \{ 0\}\ \cup S^{n-1} \times I) \cup_H X$. It is a bijection because is the identity on $X$ and maps $\mathring{D}^n \subset D^n \cup_f X$ homeomorphically onto $\mathring{D}^n \times \{ 0\} \subset (D^n \times \{ 0\}\ \cup S^{n-1} \times I) \cup_H X$. It is easy to see that $\iota'$ is a closed map,
hence a homeomorphism.
