How to show $\frac{X\sqcup_{S^1}D^2}{D^2}\simeq \frac{X}{S^1}$ Let $X$ be a $CW$-complex which contains $S^1$. How to show $X\sqcup_{S^1} D^2/D^2$ is homeomorphic to $X/S^1$? 
Here $D^2$ is two dimensional closed unit disc in $\mathbb R^2$.
My Atempt: Let $q^{'}$ be the composition $$X\hookrightarrow X\sqcup D^2\rightarrow X\sqcup_{S^1} D^2\rightarrow \frac{X\sqcup_{S^1}D^2}{D^2}.$$
If I could show $q^{'}$ is a quotient map which makes the same identifications as $q:X\longrightarrow X/S^1$ then I would find $$\frac{X\sqcup_{S^1} D^2}{D^2}\simeq \frac{X}{S^1}.$$
Question: Is it true $q^{'}$ is a quotient map which the same identifications as $q$?
 A: Note that 
$X\hookrightarrow X\sqcup D^2\to X\sqcup_{S^1} D^2$
is a closed embedding of $X$ into the adjunction space. 

In general, if $Y$ is glued to $X$ via a map $f$ from a subpace $A$
  $$\begin{array}{ccc}
A & \hookrightarrow & Y \\
\ \downarrow f & & \downarrow \\
X & \to  & Y\cup_f X
\end{array}$$
  then $X$ will be a subspace of the $Y\cup_f X$, and if $A$ is closed in $Y$, then $X$ will be a closed subspace.

$X$ is a closed subspace of $X\sqcup_{S^1} D^2$. If you take a closed and $d$-saturated subspace $C$ of this $X$, where $d:X\sqcup_{S^1} D^2\to\frac{X\sqcup_{S^1}D^2}{D^2}$, then its saturation in $X\sqcup_{S^1} D^2$ is either $C$ itself (if $C$ is disjoint from $S^1$), or is $C\cup D^2$ (if $C\supseteq S^1$), so it is closed. That means we have a quotient map $X\to\frac{X\sqcup_{S^1}D^2}{D^2}$. This map and $X\to X/S^1$ both contract the circle to a point, so the homeomorphism exists.
Edit: The proof can actually be simplified. If we take $d$  to be the quotient map $X⊔D^2→\frac{X⊔_{S^1}D^2}{D^2}$, then a $d$-saturated closed subset $C$ of $X$ has as its saturation either $C$ itself or $C⊔ D^2$, so it is the intersection of $X$ with a closed and saturated subset of $X⊔D^2$.
A: In some sense yes. To obtain $X\sqcup_{S^1} D^2/D^2$ you attach a 2-cell and then take the quotient with a 2-cell hence this procedure contracts the boundary of the embedded $2-disk$ where you glue that 2-cell namely the resulting space is $\frac{X}{S^1}$.
