Proof. Let $X$ be the $Stone$-$\breve{C}ech$ compactification $\beta \mathbb{\mathbb{N}}$ of $\mathbb{\mathbb{N}}$. Then $X$ is compact, and so it is a $k$-space. Since each convergent sequence in $\beta \mathbb{\mathbb{N}}$ is trivial, $\mathcal{P}=\left\{\left\{x\right\}:x\in X\right\}$ is an $sn$-network of $X$. It is clear that $\mathcal{P}$ is not a weak-base of $X$. □

Proof. (1) $\Rightarrow$ (2). Note that a space with a locally countable $sn$-network is $sn$-first countable. So (1) $\Rightarrow$ (2) by Remark 2.5(1) and Lemma 2.9.

(2) $\Rightarrow$ (1). By Lemma 2.9, let $\mathcal{P}$ be a locally countable $cs$-network of $X$. We can assume that $\mathcal{P}$ is closed under finite intersections. For each $x\in X$, let $\left\{B_n\left(x\right):n\in \mathbb{\mathbb{N}}\right\}$ be a countable $sn$-network at $x$ in $X$, and let ${\mathcal{P}}_x=\left\{P\in \mathcal{P}:B_n\left(x\right)\subset Pforsomen\in \mathbb{\mathbb{N}}\right\}$. Then each element of ${\mathcal{P}}_x$ is a sequential neighborhood of $X$. Put ${\mathcal{P}}^{\prime }=\bigcup \left\{{\mathcal{P}}_x:x\in X\right\}$, then ${\mathcal{P}}^{\prime }\subset \mathcal{P}$ is locally countable. It suffices to prove that ${\mathcal{P}}_x$ is a network at $x$ in $X$ for each $x\in X$. If not, there is an open neighborhood $U$ of $x$ such that $P\not\subset U$ for each $P\in {\mathcal{P}}_x$. Let $\left\{P\in \mathcal{P}:x\in P\subset U\right\}=\left\{P_m\left(x\right):m\in \mathbb{\mathbb{N}}\right\}$. Then $B_n\left(x\right)\not\subset P_m\left(x\right)$ for each $n,m\in \mathbb{\mathbb{N}}$. Choose $x_{n,m}\in B_n\left(x\right)-P_m\left(x\right)$. For $n\ge m$, let $x_{n,m}=y_k$, where $k=m+n\left(n-1\right)∕2$. Then the sequence $\left\{y_k:k\in \mathbb{\mathbb{N}}\right\}$ converges to $x$. Thus , there is $m,i\in \mathbb{\mathbb{N}}$ such that $\left\{y_k:k\ge i\right\}\bigcup \left\{x\right\}\subset P_m\left(x\right)\subset U$. Take $j\ge i$ with $y_j=x_{n,m}$ for some $n\ge m$. Then $x_{n,m}\in P_m\left(x\right)$. This is a contradiction.

(1) $\Rightarrow$ (3). Let $\mathcal{P}$ be a locally countable $sn$-network of $X$. For each $x\in X$, there is an open neighborhood $U$ of $x$ such that ${\mathcal{P}}_U=\left\{P\bigcap U:P\in \mathcal{P}\right\}$ is countable. It is easy to prove that ${\mathcal{P}}_U$ is a countable $sn$-network of subspace $U$. So $U$ is an $sn$-second countable space. Thus $X$ is a locally $sn$-second countable space.

(3) $\Rightarrow$ (4) $\Rightarrow$ (5). It is known that $sn$-second countable $\Rightarrow$ ${\aleph }_0$ $\Rightarrow$ hereditarily separable. So (3) $\Rightarrow$ (4) $\Rightarrow$ (5).

(5) $\Rightarrow$ (6). It suffices to prove that $X$ is locally hereditarily $Lindel\ddot{o}f$. Let $x\in X$ and $U$ be a hereditarily separable neighborhood of $x$. By Lemma 2.11, $U$ is hereditarily $Lindel\ddot{o}f$. So $X$ is locally hereditarily $Lindel\ddot{o}f$.

(6) $\Rightarrow$ (1). Let $\mathcal{P}=\bigcup \left\{{\mathcal{P}}_n:n\in \mathbb{\mathbb{N}}\right\}$ be a $\sigma$-locally countable $sn$-network of a Locally $Lindel\ddot{o}f$ space $X$, where each ${\mathcal{P}}_n$ is locally countable in $X$. Let $x\in X$ and let $U$ be a $Lindel\ddot{o}f$ neighborhood of $x$. Let $n\in \mathbb{\mathbb{N}}$. For each $y\in U$, there is an open neighborhood $U_y$ of $y$ such that $U_y$ intersects at most countable many elements of ${\mathcal{P}}_n$. The open cover $\left\{U_y:y\in U\right\}$ of $U$ has countable subcover $\mathcal{V}$. Put $V=\bigcup \mathcal{V}$, then $U\subset V$ and $V$ intersects at most countable many elements of ${\mathcal{P}}_n$. So $U$ intersects at most countable many elements of ${\mathcal{P}}_n$. Moreover, $U$ intersects at most countable many elements of $\mathcal{P}$. Thus $\mathcal{P}$ is a locally countable $sn$-network of $X$. □

Proof. Let $\mathcal{P}$ be a $\sigma$-locally countable $c{s}^{\ast }$-network of $X$. Whenever $K$ is a compact subset of $X$, put ${\mathcal{P}}_K=\left\{P\bigcap K:P\in \mathcal{P}\right\}$, then ${\mathcal{P}}_K$ is a $\sigma$-locally countable $c{s}^{\ast }$-network of $K$. It is easy to see that ${\mathcal{P}}_K$ is a countable $c{s}^{\ast }$-network of $K$, and so $K$ has a countable $k$-network by Remark 2.8(4). By Lemma 2.14(1), $K$ is metrizable. So $X$ has a point-countable $k$-network by Remark 2.14(3), thus $X$ is sequential by Remark 2.14(2). □

Proof. (1) $\Rightarrow$ (2). $X$ is a $k$-space with a locally countable $cs$-network, so $X$ is a topological sum of ${\aleph }_0$-spaces([17, Theorem 1]). It is easy to see that $sn$-first countability is hereditary to subspace. Note that each $sn$-first countable, ${\aleph }_0$-space is $sn$-second countable([13, Theorem 2.1]). So $X$ is a topological sum of $sn$-second countable spaces.

(2) $\Rightarrow$ (3). Let $X=\oplus \left\{X_{\alpha }:\alpha \in \Lambda \right\}$, where each $X_{\alpha }$ is $sn$-second countable. Note that each $X_{\alpha }$ is a (hereditarily) separable, open subspace of $X$, So $X$ is locally (hereditarily) separable. For each $\alpha \in \Lambda$, let $\left\{P_{\alpha ,n}:n\in \mathbb{\mathbb{N}}\right\}$ be a countable $sn$-network of $X_{\alpha }$. Put ${\mathcal{P}}_n=\left\{P_{\alpha ,n}:\alpha \in \Lambda \right\}$ for each $n\in \mathbb{\mathbb{N}}$, and put $\mathcal{P}=\bigcup \left\{{\mathcal{P}}_n:n\in \mathbb{\mathbb{N}}\right\}$, then $\mathcal{P}$ is a $\sigma$-locally finite $sn$-network of $X$. So $X$ is an $sn$-metrizable space.

(3) $\Rightarrow$ (4). It is clear.

(4) $\Rightarrow$ (1). By Remark 2.13, it suffices to prove that $X$ is locally $Lindel\ddot{o}f$. Let $\mathcal{P}$ be a $\sigma$-locally countable $sn$-network of $X$. $X$ is a sequential space by Lemma 2.15, so $\mathcal{P}$ is a $\sigma$-locally countable $k$-network of $X$([30, Corollary 1.5]). Recalled a space is meta-$Lindel\ddot{o}f$ if each open cover of it has a point countable open refinement. Thus $X$ is hereditarily meta-$Lindel\ddot{o}f$([17, Proposition 1]). Each hereditarily meta-$Lindel\ddot{o}f$, locally separable space is locally $Lindel\ddot{o}f$([14, Proposition 8.7]), so $X$ is locally $Lindel\ddot{o}f$. □

Proof. Let $D$ is a discrete space, where $\mid D\mid =2^{\omega }$. By [3, Example 4.2], there is an almost disjoint family $\left\{{\mathcal{P}}_{\alpha }:\alpha <2^{\omega }\right\}$ consisting of countable infinite subsets of $D$ such that for each uncountable subset $P$ of $D$, there is $\alpha <2^{\omega }$ such that $P_{\alpha }\subset P$. Let $\left\{P_{\alpha ,n}:n\in \mathbb{\mathbb{N}}\right\}$ be a mutually disjoint family consisting of infinite subsets of $P_{\alpha }$. For each $\alpha <2^{\omega }$ and each $n\in \mathbb{\mathbb{N}}$, choose $p_{\alpha ,n}\in \overline{P_{\alpha ,n}}-P_{\alpha ,n}$, where $\overline{P_{\alpha ,n}}$ is the closure of $P_{\alpha ,n}$ in the $Stone$-$\breve{C}ech$ compactification $\beta D$ of $D$. Put $X=D\bigcup \left\{p_{\alpha ,n}:\alpha <2^{\omega },n\in \mathbb{\mathbb{N}}\right\}$, and $X$ is endowed the subspace topology of $\beta D$.

Claim 1. $X$ has a $\sigma$-locally countable $sn$-network.

By [22, Example 5.1.18(1)], $X$ has a $\sigma$-locally countable $cs$-network $\mathcal{P}$. Note that each compact subset of $X$ is finite [22, Example 1.5.5], so each convergent sequence of $X$ is finite. Thus we can assume that $\mathcal{P}$ is closed under finite intersections. It is easy to see that $\mathcal{P}$ is an $sn$-network of $X$. So $X$ has a $\sigma$-locally countable $sn$-network.

Claim 2. $X$ is not a topological sum of ${\aleph }_0$-spaces [22, Example 5.1.18(1)]. □

Proof. Let $X={\omega }_1\bigcup \left({\omega }_1\times \left\{1∕n:n\in \mathbb{\mathbb{N}}\right\}\right)$. Define a neighborhood base ${\mathcal{ℬ}}_x$ for each $x\in X$ for the desired topology on $X$ as follows.

(1) If $x\in X-{\omega }_1$, then ${\mathcal{ℬ}}_x=\left\{\left\{x\right\}\right\}$.

(2) If $x\in {\omega }_1$, then ${\mathcal{ℬ}}_x=\left\{\left\{x\right\}\bigcup \left(\bigcup \left\{V\left(n,x\right)\times \left\{1∕n\right\}:n\ge m\right\}\right):m\in \mathbb{\mathbb{N}}andV\left(n,x\right)isaneighborhoodofxin{\omega }_{1}withtheordertopology\right\}$.

By [17, Example 1], $X$ has a locally countable $k$-network, which is not an $\aleph$-space. It suffices to prove that $X$ is $sn$-first countable by Remark 2.13.

Let $x\in X$. If $x\in X-{\omega }_1$, then $\left\{\left\{x\right\}\right\}$ is a countable $sn$-network at $x$ in $X$. If $x\in {\omega }_1$, put ${\mathcal{P}}_x=\left\{P_{x,m}:m\in \mathbb{\mathbb{N}}\right\}$, where $P_{x,m}=\left\{x\right\}\bigcup \left\{\left(x,1∕n\right):n\ge m\right\}$. Then ${\mathcal{P}}_x$ is a countable network at $x$ in $X$. We only need to prove that each $P_{x,m}$ is a sequential neighborhood of $x$.

Let $\left\{x_n\right\}$ be a sequence converging to $x$. Put $K=\left\{x\right\}\bigcup \left\{x_n:n\in \mathbb{\mathbb{N}}\right\}$, then $K$ is a compact subset of $X$. By [17, Example 1], we have the following facts.

Fact 1. $K\bigcap {\omega }_1$ is finite.

Fact 2. $K-\bigcup \left\{\left\{y\right\}\bigcup \left\{\left(y,1∕n\right):n\in \mathbb{\mathbb{N}}\right\}:y\in K\bigcap {\omega }_1\right\}$ is finite.

Case 1. If there is $y\in K\bigcap {\omega }_1$ such that $y=x_n$ for infinite many $n\in \mathbb{\mathbb{N}}$, i.e., there is a subsequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ such that $y=x_{n_k}$ for each $k\in \mathbb{\mathbb{N}}$, then $y=x$, So $\left\{x_n\right\}$ is frequently in $P_{x,m}$.

Case 2. If Case 1 does not hold, without loss of the generalization, we may assume $K\bigcap {\omega }_1=\left\{x\right\}$ by Fact 1. By Fact 2, $K-\left\{x\right\}\bigcup \left\{\left(x,1∕n\right):n\in \mathbb{\mathbb{N}}\right\}$ is finite. If there is $y\in K-\left\{x\right\}\bigcup \left\{\left(x,1∕n\right):n\in \mathbb{\mathbb{N}}\right\}$ such that $y=x_n$ for infinite many $n\in \mathbb{\mathbb{N}}$, then $\left\{x_n\right\}$ is frequently in $P_{x,m}$ by a similar way in the proof of Case 1. Conversely, there is $k_0\in \mathbb{\mathbb{N}}$ such that $\left\{x\right\}\bigcup \left\{x_n:n\ge k_0\right\}\subset \left\{x\right\}\bigcup \left\{\left(x,1∕n\right):n\in \mathbb{\mathbb{N}}\right\}$. So $\left\{x_n\right\}$ is eventually in $P_{x,m}$.

By the above Case 1 and Case 2, $P_{x,m}$ is a sequential neighborhood of $x$ by Remark 2.2(1). □

Proof. Let $D$ be a sequentially dense subset of $X$, and let $\mathcal{P}=\left\{{\mathcal{P}}_x:x\in X\right\}$, where ${\mathcal{P}}_x$ is an $sn$-network at $x$ in $X$ for each $x\in X$. For each $x\in D$, ${\left(\mathcal{P}\right)}_x$ is countable because $\mathcal{P}$ is point countable, where ${\left(\mathcal{P}\right)}_x=\left\{P\in \mathcal{P}:x\in P\right\}$. Furthermore, $\bigcup \left\{{\left(\mathcal{P}\right)}_x:x\in D\right\}$ is countable. For each $x\in X$ and $P\in {\mathcal{P}}_x$, there is a sequence $S$ in $D$ converging to $x$. Note that $P$ is a sequential neighborhood of $x$. $S$ is eventually in $P$, and so $P\bigcap D\ne \varnothing$. This proves that each element of $\mathcal{P}$ intersects with $D$, thus $\mathcal{P}=\bigcup \left\{{\left(\mathcal{P}\right)}_x:x\in D\right\}$. So $\mathcal{P}$ is countable. □

Proof. Since $\sigma$-locally countable $\Rightarrow$ point countable, we only need to prove parenthetic part.

Let $X$ be locally sequentially separable. For each $x\in X$, there is an open neighborhood $U$ of $x$ such that $U$ is sequentially separable. It is clear that $\left\{P\bigcap U:P\in \mathcal{P}\right\}$ is a point countable $sn$-network of $U$. $\left\{P\bigcap U:P\in \mathcal{P}\right\}$ is countable by Proposition 2.20, So $\mathcal{P}$ is locally countable in $X$. □

Proof. Let $\mathbb{\mathbb{Q}}\subset X\subset \mathbb{ℝ}$ and $\mid X\mid >\omega$, where $\mathbb{\mathbb{Q}}$ and $\mathbb{ℝ}$ are the set of all rational numbers and the set of all real numbers respectively. Let $Y=X\bigcup \left(\bigcup \left\{\mathbb{\mathbb{Q}}\times \left\{1∕n\right\}:n\in \mathbb{\mathbb{N}}\right\}\right)$. Define a neighborhood base ${\mathcal{ℬ}}_y$ for each $y\in Y$ for the desired topology on $Y$ as follows.

(1) If $y\in Y-X$, then ${\mathcal{ℬ}}_y=\left\{\left\{y\right\}\right\}$.

(2) If $y\in X$, then ${\mathcal{ℬ}}_y=\left\{\left\{y\right\}\bigcup \left(\bigcup \left\{\left(\left[a_{y,n},y\right)\bigcap \mathbb{\mathbb{Q}}\right)\times \left\{1∕m\right\}:n\ge m\right\}\right):m\in \mathbb{\mathbb{N}}andy>a_{y,n}\in \mathbb{ℝ}\right\}$.

Then $Y$ is a separable, $\aleph$-space and not an ${\aleph }_0$-space [16, Example 1]. On the other hand, each compact subset of $Y$ is finite [16, Example 1]. By a similar way as in the proof of Example 2.18(claim 1), we can prove $Y$ has a $\sigma$-locally finite $sn$-network. That is, $Y$ is an $sn$-metrizable space. □

Proof. If $Y$ has a locally countable $sn$-network, then $X$ has a locally countable $cs$-network by Remark 2.5(1), Lemma 2.9 and Lemma 3.6. We only need to prove that $X$ is $sn$-first countable by Remark 2.13. Since $X$ has a $G_{\delta }$-diagonal, $X$ is point-$G_{\delta }$. By Lemma 3.5, It suffices to prove that $X$ contains no closed subspace having $S_{\omega }$ as its sequential coreflection.

Assume $X$ contains closed subspace $S$ having $S_{\omega }$ as its sequential coreflection. Put $g=f{\mid }_{\sigma s}:\sigma S\to \sigma f\left(S\right)$.

Claim 1. $g$ is closed.

Proof. Let $A$ be a closed subset of $\sigma S$, then $A$ is sequentially closed in $S$. It is clear that $f:S\to f\left(S\right)$ is closed and $S$ is point-$G_{\delta }$. So $f\left(A\right)$ is sequentially closed in $f\left(S\right)$ by Lemma 3.7, thus $f\left(A\right)$ is closed in $\sigma f\left(S\right)$.

Claim 2. $g^{-1}\left(y\right)$ is compact in $\sigma S$ for each $y\in \sigma f\left(S\right)$.

Proof. Let $y\in \sigma f\left(S\right)$. Note that $X$ has a $G_{\delta }$-diagonal and $f^{-1}\left(y\right)$ is compact in $X$, so $f^{-1}\left(y\right)$ is metrizable [5]. Therefore, the topology on the sequential coreflection of $f^{-1}\left(y\right)\bigcap S$ is equivalent to the induced topology of subspace $S$ of $X$. Thus $g^{-1}\left(y\right)=f^{-1}\left(y\right)\bigcap S$ is compact in $\sigma S$.

By the above two claims, $g$ is perfect. Since $S_{\omega }$, which is homeomorphic to $\sigma S$, is a $Fréchet$, $\aleph$-space and perfect mappings preserve $Fréchet$, $\aleph$-spaces, $\sigma f\left(S\right)$ is a $Fréchet$, $\aleph$-space. On the other hand, $Y$ is $sn$-first countable, so $f\left(S\right)$, as a subspace of $Y$, is $sn$-first countable. By [20, Theorem 3.13], $\sigma f\left(S\right)$ is $g$-first countable, so $\sigma f\left(S\right)$ is metrizable [11, Theorem 2.4], and so $\sigma S$ is metrizable [5]. This contradicts that $S_{\omega }$ is not metrizable. □

Proof. Let $X=\left\{0\right\}\bigcup \mathbb{\mathbb{N}}\bigcup \left(\mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}\right)$, $\mathcal{ℱ}=\left\{F\subset \mathbb{\mathbb{N}}:Fisfinite\right\}$, ${\mathbb{\mathbb{N}}}^{\mathbb{\mathbb{N}}}=\left\{f:f\text{is a correspondence from }\mathbb{\mathbb{N}}to\mathbb{\mathbb{N}}\right\}$. For $n,m,k\in \mathbb{\mathbb{N}}$, $F\in \mathcal{ℱ}$, and $f\in {\mathbb{\mathbb{N}}}^{\mathbb{\mathbb{N}}}$, put $V\left(n,m\right)=\left\{n\right\}\bigcup \left\{\left(n,k\right):k\ge m\right\}$, $H\left(F,f\right)=\bigcup \left\{V\left(n,f\left(n\right)\right):n\in \mathbb{\mathbb{N}}-F\right\}$. Define a neighborhood base ${\mathcal{ℬ}}_x$ for each $x\in X$ for the desired topology on $X$ as follows.

(1) If $x\in \mathbb{\mathbb{N}}\times \mathbb{\mathbb{N}}$, then ${\mathcal{ℬ}}_x=\left\{\left\{x\right\}\right\}$.

(2) If $x\in \mathbb{\mathbb{N}}$, then ${\mathcal{ℬ}}_x=\left\{V\left(x,m\right):m\in \mathbb{\mathbb{N}}\right\}$.

(3) If $x=0$, then ${\mathcal{ℬ}}_x=\left\{\left\{x\right\}\bigcup H\left(F,f\right):F\in \mathcal{ℱ},f\in {\mathbb{\mathbb{N}}}^{\mathbb{\mathbb{N}}}\right\}$.

Let $Y$ be the quotient space obtained from $X$ by shrinking the set $\left\{0\right\}\bigcup \mathbb{\mathbb{N}}$ to a point, $f:X\to Y$ be a natural mapping. Then

Claim 1. $f$ is perfect and $X$ is $g$-second countable [18, Example 3.1].

Claim 2. $Y$ is not $sn$-first countable [11, Example 3.2], so $Y$ has not any locally countable $sn$-network. □

Proof. Let $y\in Y$. Then there is a finite subset $F$ of $f^{-1}\left(y\right)$, such that for any sequence $S$ in $Y$ converging to $y$, there is a sequence $L$ in $X$ converging to some $x\in F$ and $f\left(L\right)$ is a subsequence of $S$. $X$ is $sn$-first countable, for each $x\in F$, let ${\mathcal{P}}_x=\left\{P_{x,n}:n\in \mathbb{\mathbb{N}}\right\}$ be a decreasing $sn$-network at $x$ in $X$. Put ${\mathcal{ℱ}}_y=\left\{\bigcup \left\{f\left(P_{x,n}\right):x\in F\right\}:n\in \mathbb{\mathbb{N}}\right\}$. Then ${\mathcal{ℱ}}_y$ is countable decreasing.

(1) ${\mathcal{ℱ}}_y$ is a network at $y$ in $Y$. In fact, let $U$ be an open neighborhood of $y$, then $F\subset f^{-1}\left(y\right)\subset f^{-1}\left(U\right)$. For each $x\in F$, there is $n_x\in \mathbb{\mathbb{N}}$ such that $x\in P_{x,n_x}\subset f^{-1}\left(U\right)$, so $y\in f\left(P_{x,n_x}\right)\subset U$. Put $n_0=max\left\{n_x:x\in F\right\}$, then $P_{x,n_0}\subset P_{x,n_x}$ for each $x\in F$. So $y\in \bigcup \left\{f\left(P_{x,n_0}\right):x\in F\right\}\subset \bigcup \left\{f\left(P_{x,n_x}\right):x\in F\right\}\subset U$.

(2) Let $\bigcup \left\{f\left(P_{x,n_1}\right):x\in F\right\},\bigcup \left\{f\left(P_{x,n_2}\right):x\in F\right\}\in {\mathcal{ℱ}}_y$. Put $n_0=max\left\{n_1,n_2\right\}$, then $\bigcup \left\{f\left(P_{x,n_0}\right):x\in F\right\}\in {\mathcal{ℱ}}_y$ and $\bigcup \left\{f\left(P_{x,n_0}\right):x\in F\right\}\subset \left(\bigcup \left\{f\left(P_{x,n_1}\right):x\in F\right\}\right)\bigcap \left(\bigcup \left\{f\left(P_{x,n_2}\right):x\in F\right\}\right)$.

(3) $\bigcup \left\{f\left(P_{x,n}\right):x\in F\right\}$ is a sequential neighborhood of $y$ for each $n\in \mathbb{\mathbb{N}}$. In fact, let $S$ be a sequence in $Y$ converging to $y$. Then there is a sequence $L$ in $X$ converging to some $x_0\in F$ and $f\left(L\right)$ is a subsequence of $S$. For each $n\in \mathbb{\mathbb{N}}$. Since $P_{x_0,n}$ is a sequential neighborhood of $x$, $L$ is eventually in $P_{x_0,n}$. So $f\left(L\right)$ is eventually in $f\left(P_{x_0,n}\right)$, hence $S$ is frequently in $f\left(P_{x_0,n}\right)$. Moreover, $S$ is frequently in $\bigcup \left\{f\left(P_{x,n}\right):x\in F\right\}$. By Remark 2.2(1), $\bigcup \left\{f\left(P_{x,n}\right):x\in F\right\}$ is a sequential neighborhood of $y$. □

Proof. Let $\mathcal{P}=\left\{P_{\alpha }:\alpha \in \Lambda \right\}$ be a locally countable family of subsets of $X$ and let $y\in Y$. For each $x\in f^{-1}\left(y\right)$, there is an open neighborhood $U_x$ of $x$ such that $\left\{\alpha \in \Lambda :U_x\bigcap P_{\alpha }\ne \varnothing \right\}$ is countable. $f^{-1}\left(y\right)\subset \bigcup \left\{U_x:x\in f^{-1}\left(y\right)\right\}$ and $f^{-1}\left(y\right)$ is $Lindel\ddot{o}f$, so there is a countable subset $B$ of $f^{-1}\left(y\right)$ such that $f^{-1}\left(y\right)\subset \bigcup \left\{U_x:x\in B\right\}$. Put $U=\bigcup \left\{U_x:x\in B\right\}$. It is clear that $\left\{\alpha \in \Lambda :U\bigcap P_{\alpha }\ne \varnothing \right\}$ is countable. Note that $f$ is closed. By [7, Theorem 1.4.13], there is an open neighborhood $V$ of $y$ such that $f^{-1}\left(V\right)\subset U$. Thus ${\Lambda }^{\prime }=\left\{\alpha \in \Lambda :f^{-1}\left(V\right)\bigcap P_{\alpha }\ne \varnothing \right\}$ is countable. It is easy to check that $\left\{\alpha \in \Lambda :V\bigcap f\left(P_{\alpha }\right)\ne \varnothing \right\}={\Lambda }^{\prime }$. So $\left\{\alpha \in \Lambda :V\bigcap f\left(P_{\alpha }\right)\ne \varnothing \right\}$ is countable. This proves that $f\left(\mathcal{P}\right)$ is a locally countable family of subsets of $Y$. □

Proof. Let $\mathcal{P}$ be a locally countable $sn$-network of $X$. Then $f$ is sequentially quotient by Remark 3.2(3), and so $Y$ is $sn$-first countable by Remark 3.2(1) and Lemma 3.11. Since sequentially quotient mappings preserve $c{s}^{\ast }$-networks [19, Proposition 2.7.3], $f\left(\mathcal{P}\right)$ is a $c{s}^{\ast }$-network of $Y$. $f\left(\mathcal{P}\right)$ is locally countable by Lemma 3.12, so $f\left(\mathcal{P}\right)$ is a locally countable $c{s}^{\ast }$-network of $Y$. Thus $Y$ has a locally countable $sn$-network by Remark 2.13. □

Proof. Let $\mathcal{P}=\bigcup \left\{{\mathcal{P}}_x:x\in X\right\}$ be a locally countable $sn$-network of $X$. Since $f$ is closed, $Lindel\ddot{o}f$, by a similar way as in the proof of Theorem 3.13, $f\left(\mathcal{P}\right)$ is a locally countable $c{s}^{\ast }$-network of $Y$. It suffices to prove that $Y$ is $sn$-first countable by Remark 2.13. Let $y\in Y$. Put ${\mathcal{ℱ}}_y=\left\{f\left(P\right):P\in {\mathcal{P}}_{x}andx\in f^{-1}\left(y\right)\right\}$, then ${\mathcal{ℱ}}_y\subset f\left(\mathcal{P}\right)$, so ${\mathcal{ℱ}}_y$ is locally countable. Note that $y\in \bigcap {\mathcal{ℱ}}_y$, ${\mathcal{ℱ}}_y$ is countable. It is clear that ${\mathcal{ℱ}}_y$ is a network at $y$ in $Y$. We only need to prove that each element of ${\mathcal{ℱ}}_y$ is a sequential neighborhood of $y$. Let $f\left(P\right)\in {\mathcal{ℱ}}_y$ and $\left\{y_k\right\}$ be a sequence in $Y$ converging to $y$. Then there is $x\in f^{-1}\left(y\right)$ such that $P\in {\mathcal{P}}_x$. Since $X$ is point-$G_{\delta }$, $\left\{x\right\}=\bigcap \left\{U_n:n\in \mathbb{\mathbb{N}}\right\}$, where each $U_n$ is open in $X$ and $\overline{U_{n+1}}\subset U_n$. For each $n\in \mathbb{\mathbb{N}}$, $y\in f\left(U_n\right)$ and $f\left(U_n\right)$ is open as $f$ is open, so there is $m_n\in \mathbb{\mathbb{N}}$ such that $y_k\in f\left(U_n\right)$ for each $k\ge m_n$. Pick $x_n\in U_n$ such that $f\left(x_n\right)=y_{m_n}$. Since $f$ is closed, it is not difficult to prove that the sequence $\left\{x_n\right\}$ converges to $x\in P$. $P$ is a sequential neighborhood of $x$, so $\left\{x_n\right\}$ is eventually in $P$. Consequently, $\left\{f\left(x_n\right)\right\}$ is eventually in $f\left(P\right)$, so $\left\{y_k\right\}$ is frequently in $f\left(P\right)$. By Remark 2.2(1), $f\left(P\right)$ is a sequential neighborhood of $y$. □

Proof. Let $\mathcal{P}=\left\{P_{\alpha }:\alpha \in \Lambda \right\}$ be a locally countable family of subsets of $X$ and let $y\in Y$. Then there is a neighborhood $W$ of $y$ in $Y$ such that $f^{-1}\left(\overline{W}\right)$ is a $Lindel\ddot{o}f$ subset of $X$. It suffices to prove that $\left\{\alpha \in \Lambda :W\bigcap f\left(P_{\alpha }\right)\ne \varnothing \right\}$ is countable. For each $x\in f^{-1}\left(\overline{W}\right)$, there is an open neighborhood $U_x$ of $x$ such that $\left\{\alpha \in \Lambda :U_x\bigcap P_{\alpha }\ne \varnothing \right\}$ is countable. $f^{-1}\left(\overline{W}\right)\subset \bigcup \left\{U_x:x\in f^{-1}\left(\overline{W}\right)\right\}$ and $f^{-1}\left(\overline{W}\right)$ is $Lindel\ddot{o}f$, so there is a countable subset $B$ of $f^{-1}\left(\overline{W}\right)$ such that $f^{-1}\left(\overline{W}\right)\subset \bigcup \left\{U_x:x\in B\right\}$. It is easy to see that $\left\{\alpha \in \Lambda :\left(\bigcup \left\{U_x:x\in B\right\}\right)\bigcap P_{\alpha }\ne \varnothing \right\}$ is countable, so ${\Lambda }^{\prime }=\left\{\alpha \in \Lambda :f^{-1}\left(W\right)\bigcap P_{\alpha }\ne \varnothing \right\}$ is countable. It is easy to check that $\left\{\alpha \in \Lambda :W\bigcap f\left(P_{\alpha }\right)\ne \varnothing \right\}={\Lambda }^{\prime }$. This completes the proof. □

Proof. We only need to prove part (1) by Remark 3.2(1). Let $f:X\to Y$ be a finite subsequence-covering, strongly $Lindel\ddot{o}f$ mapping and $\mathcal{P}$ be a locally countable $sn$-network of $X$. Then $Y$ is $sn$-first countable by lemma 3.11 and $f\left(\mathcal{P}\right)$ is a locally countable family of subsets of $Y$ by Lemma 3.18. By a similar way as in the proof of Theorem 3.13, we can prove $f\left(\mathcal{P}\right)$ is a $c{s}^{\ast }$-network of $Y$. So $Y$ has a locally countable $sn$-network by Remark 2.13. □