How to show $\frac{X\sqcup_{S^1}D^2}{D^2}\simeq \frac{X}{S^1}$

Question

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 $q$?

Answer 1

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 $X⊔D^2$.

Answer 2

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 $\frac{X}{S^1}$.