Adjunction spaces of second countable, locally compact, Hausdorff spaces

Question

Suppose that $X$ and $Y$ are second countable, locally compact, Hausdorff spaces. Let $Z$ be a closed subspace of $X$ and suppose that $f : Z \to Y$ is an injective and open continuous map (hence a homeomorphism onto it's image). We can form the adjunction space, $$ X \sqcup_f Y = (X \sqcup Y) / \{z \sim f(z) \mid z \in Z\}.$$

Is $X \sqcup_f Y$ nessesarily second countable, locally compact, and Hausdorff?

---

So far I can show that since $X$ and $Y$ are normal then $X \sqcup_f Y$ is normal (an argument can be found in Lemma 1 here, it holds without my assumptions on $f$). Second countablility and local compactness have me stumped though.

Answer 1

No. Let $Y=\{0\}\cup\{1/n: n\in\Bbb N\}$ be a sequence convergent to zero, $X=Y\times\Bbb N$, $Z=\{0\}\times\Bbb N$, and $f:Z\to Y$, $(0,n)\mapsto 1/n$. It is easy to see that the space $X \sqcup_f Y$ is homeomorphic to the following Franklin’s space decribed in the following example from “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989).

Moreover, let $q: X \sqcup Y\to X \sqcup_f Y$ be the quotient map. It is easy to see that any neighborhood of $q(0)$ is $X \sqcup_f Y$ contains a sequence of a form $\{q(n, m_n):n\ge N\}$ for some $N$, which has no limit points, so $X \sqcup_f Y$ is not locally compact.

Answer 2

(For simplicity, I identify $X$ and $Y$ with their canonical images in $X\sqcup_f Y$).

Second counatbility: Let $\{U_i\}_{i\in\Bbb N}$ be a countable base of $X$. For each $i$, $U_i\cap Z$ is open in $Z$, hence its image under $f$ is open in $f(Z)$, i.e., there exists $U_i'$ open in $Y$ such that $f(U_i\cap Z)=U_i'\cap f(Z)$. Together, $U_i$ and $U_i'$ form an open set $U''_i\subseteq X\sqcup_f$ such that $U_i''\cap Z/f=Z/f$. Likewise, from a countable base $\{V_i\}_{i\in\Bbb N}$, we find $V_i'$ open in $X$ such that $f(V_i'\cap Z)=V_i\cap f(Z)$ and with that $V''_i$. Then the following together make up a countable base of $X\sqcup_f Y$:

• all $U_i\setminus Z$
• all $V_i\setminus f(Z)$
• all $(U''_i\cap V''_j)$

To see this, consider any open $W$ in $X\sqcup_f Y$. Then $W\cap X=\bigcup_{i\in I} U_i$ and $W\cap Y=\bigcup_{j\in J}V_j$ for suitable subsets $I,J\subseteq \Bbb N$. Now verify that $$W=\bigcup _{i\in I}(U_i\setminus Z)\cup \bigcup _{j\in J}(V_j\setminus f(Z))\cup \bigcup_{(i,j)\in I\times J}(U''_i\cap V''_j)$$