Proving $X^{n-1} \cong (X^{n-1} \sqcup_\alpha \partial D^n_\alpha) / (x \sim \varphi_\alpha(x))$ in CW-complexes

Question

In this answer, it is mentioned that $$X^{n-1} \cong (X^{n-1} \sqcup_\alpha \partial D^n_\alpha) / (x \sim \varphi_\alpha(x))$$ where $\varphi_\alpha: \partial D^n_\alpha \rightarrow X^{n-1}$ is the attaching map.

To show this, i.e. that $X^{n-1}$ is homeomorphic to $X^{n-1} = (X^{n-1} \sqcup_\alpha \partial D^n_\alpha) / (x \sim \varphi_\alpha(x))$, we can use the result of induced homeomorphism by quotient map. This result is detailed in Corollary 22.3 of Munkres's book on Topology or here.

Specifically, it needs to be shown that $f: X^{n-1}\sqcup_\alpha \partial D^n_\alpha \rightarrow X^{n-1}$ defined as $$f(x) = \begin{cases} x_1 & (x_2 = 0)\\ \varphi_\alpha(x_1) & (x_2 = \alpha) \end{cases}$$ is a quotient map.

• Starting with continuity, this follows from the pasting lemma. To see this, note that $\varphi_\alpha$ is continuous and $(X^{n-1}, 0)$ as well as $(D^n_\alpha, \alpha)$ are open by definition of the disjoint union topology.
• Continuing with surjectivity, this is trivial as all of $X^{n-1}$ are mapped to $X^{n-1}$.
• Finally, with proving that f should be an open map, I am completely lost. This is because I don't see how it is possibly an open map because $\varphi_{\alpha}$ is not necessarily open.

Could I please get some help?

Answer 1

Let's prove that $f$ is quotient map $X^{n-1} = (X^{n-1} ⊔_{\alpha} \partial D_{\alpha}^{n}) / (x ≡ \phi_{\alpha}(x))$

This expression states that the $n-1$ dimensional cell complex $X^{n-1}$ is homeomorphic to the quotient space obtained by gluing the boundary of an n-dimensional disk $\partial D_{\alpha}^{n} \to X^{n-1}$ along the attaching map $\phi_{\alpha}$.

We will show that the map $f : X^{n-1} ⊔_{\alpha} \partial D_{\alpha}^{n} → X^{n-1}$ defined as $f(x)$ $:=$ {$x_1$ if $x_2 = 0$, $\phi_\alpha = x_1$ if $x_2 = \alpha$}.

is a quotient map. This will imply that the two spaces are homeomorphic. The map $f$ is continuous by the pasting lemma. This is because

• The attaching map $\phi_{\alpha}$ is continuous by definition.
• The spaces ($X^{n-1}, 0$) and $(D_{\alpha}^{n}, \alpha)$ are open in the disjoint union topology. To show that $f$ is a quotient map, we need to show that a subset $U ⊆ X^{n-1}$ is open if and only if its preimage $f^{-1}(U)$ is open in $X^{n-1} ⊔_{\alpha} \partial D_{\alpha}^{n}$.

If $U ⊆ X^{n-1}$ is open, then $f^{-1}(U) = U ⊔ \phi_{\alpha}^{-1}(U)$ is open in $X^{n-1} ⊔_{\alpha}\partial D_{\alpha}^{n}$ because $\phi_{\alpha}^{-1}(U)$ is open in \partial $D_{\alpha}^{n}$. Sufficiency: Conversely, suppose $f^{-1}(U)$ is open in $X^{n-1} ⊔_{\alpha} \partial D_{\alpha}^{n}$. Then $U$ is open in $X^{n-1}$ because $f^{-1}(U) \cap X^{n-1} = U$. Therefore, f is a quotient map.

$f$ is surjective because, $f$ maps $X^{n-1}$ onto itself. And f maps $D_{\alpha}^n$ onto the attached n-cell in X^n. Therefore, $f$ is surjective. Show that $f$ is injective: $f$ is injective on $X^{n-1}$ and $D_{\alpha}^n$ separately. The only points that could potentially map to the same point in X^n are those identified by the equivalence relation $x$ ~ $\phi_{\alpha}(x)$.

• But f maps these points to the same point in $X^n$, so $f$ is injective.