The excercise wants us to show that for CW complexes $X$ and $Y$ with $0$-cells $x_0$ and $y_0$, we have a homeomorphism between the two spaces $$X*Y/(X*\{y_0\} \cup \{x_0\}*Y) \cong S(X \wedge Y)/S(\{x_0\} \wedge \{y_0\}).$$ I was reading the proof in this post here as well as the solution found here. Both began with looking at the subspace $$I \times (X \times \{y_0\} \cup \{x_0\} \times Y) = (I \times X \times \{y_0\}) \cup (I \times \{x_0\} \times Y) \subset I \times X \times Y$$ and deduced that it gets identified to the subspace $X*\{y_0\} \cup \{x_0\}*Y$ in $X*Y$ which is completely fine. Then both posts claims that we automatically get a homeomorphism \begin{equation}\label{eq1}\frac{I \times X \times Y}{I \times (X \times \{y_0\} \cup \{x_0\} \times Y)} \cong \frac{X*Y}{X*\{y_0\} \cup \{x_0\}*Y}.\end{equation} My question is, why do we have this? I am assuming the quotients just mean the subspaces $I \times (X \times \{y_0\} \cup \{x_0\} \times Y)$ and $X*\{y_0\} \cup \{x_0\}*Y$ are being collapsed to a point in the respective spaces.
However, what about points of the form $(0,x_1,y_1) \in I \times X \times Y$ for $x_1 \neq x_0$ and $y_1 \neq y_1$? When constructing $X*Y$, such point is identified with $(0, x_1, y_0)$ since $\{0\} \times X \times Y$ gets collapsed to $\{0\} \times X$ in $X*Y$. However, in the above homemorphism, since such a point $(0,x_1,y_1) \notin I \times (X \times \{y_0\} \cup \{x_0\} \times Y)$ i.e. it is not identified with anything on the LHS after taking the quotient whereas, it is identified in the RHS when constructing $X*Y$. So how can the two quotient spaces be homeomorphic?