Is there some characterization of $\omega^\omega$-base related to $S_\omega$? For a topological space $X$ and one point $x\in X$, we call the cofinal type of neighborhood bases of $x$ are cofinally finer than $\omega^\omega$-base if for any neighborhood base $\mathfrak{N}$ of $x$ there exists a convergent map $f:\mathfrak{N}\rightarrow \omega^\omega$ ,i.e.,for any $\alpha\in \omega^\omega$ there exists $A\in \mathfrak{N}$ such that $\alpha\leq f(B)$ for any $B\subset A$.
If the cofinal type of neighborhood bases of $x$ are cofinally finer than $\omega^\omega$-base, does there exists an embedding $h: S_\omega\rightarrow (X,x)$ with $h(\infty)=x$? Where $S_\omega$ is the sequential fan and $\infty$ is the unique non-isolated point in $S_\omega$.
Thank you in advance.
 A: This question has negative answer.
To construct a counterexample, consider the set $X=(\omega\times\omega)\cup\{\infty\}$ where $\infty\notin(\omega\times\omega)$ is any point. Then set $X$ is endowed with the topology $\tau$ consisting of the sets $U\subset X$ satisfying the following condition:
$\bullet$ if $\infty\in U$, then there exist $n\in\omega$ and $\alpha\in\omega^\omega$ such that $\{(x,y)\in\omega\times\omega:x\ge n,\;y\ge \alpha(x)\}\subset U$.
It is easy to see that the topological space $(X,\tau)$ contains no non-trivial convergent sequences. Consequently, $(X,\tau)$ contains no copies of the sequential fan $S_\omega$.
Let us show that the cofinal type of $(X,\tau)$ at $\infty$ is finer than $\omega^\omega$.
Given any neighborhood base $\mathfrak N$ at $\infty$, for each neighborhood $U\in\mathfrak N$ of $\infty$, find $n_U\in\omega$ and a function $\alpha_U\in\omega^\omega$ such that $$\{(x,y)\in\omega\times\omega:x\ge n_U,\; y\ge \alpha_U(x-n_U)\}\subset U.$$
We claim that the function $f:\mathfrak N\to\omega^\omega$, $f:U\mapsto\alpha_U$, is convergent.
Given any function $\alpha\in\omega^\omega$, find an increasing function $\beta\in\omega^\omega$ such that $\alpha\le\beta$. Next, find a neighborhood $A\in\mathfrak N$ such that $A\subset \{(x,y)\in\omega\times\omega:y\ge \beta(x)\}$. We claim that $\alpha\le f(B)$ for any $B\in\mathfrak N$ with $B\subset A$. It follows that $\{(x,y)\in\omega\times\omega:x\ge n_B,\;y\ge \alpha_B(x-n_B)\}\subset B\subset A$.
Then for every $i\in\omega$ we have $(i+n_B,\alpha_B(i))\in B\subset A$ and hence $\alpha_B(i)\ge \beta(i+n_B)\ge \beta(i)\ge\alpha(i)$, witnessing that $f(B)=\alpha_B\ge\alpha$.
