A monotone countably cofinal function from $\omega^\omega$ to $\omega^{\omega_1}$ For a set $X$ we endow the set $\omega^X$ of all functions from $X$ to $\omega$ with the natural partial order $\le$ defined by $f\le g$ iff $f(x)\le g(x)$ for all $x\in X$.
A function $f:\omega^\omega\to\omega^X$ is called monotone if for any $\alpha\le\beta$ in $\omega^\omega$ we have $f(\alpha)\le f(\beta)$ in $\omega^X$.

Question. Is there a monotone function $f:\omega^\omega\to\omega^{\omega_1}$ such that for every countable set $A\subset\omega_1$ and every $\alpha\in\omega^A$ there exists $\beta\in\omega^\omega$ such that $\alpha\le f(\beta){\restriction}_A$?  

Remark. By Proposition 2.4.1(2) in this paper, for every monotone function $f:\omega^\omega\to\omega^{\omega_1}$ there exists $\alpha\in\omega^{\omega_1}$ such that for every $\beta\in\omega^\omega$ we have $\alpha\not\le f(\beta)$. 
PS. I learned this question from Jerzy Kąkol who arrived to it studying $\mathfrak G$-bases in locally convex spaces.
 A: The answer is no, if you require that the monotone function $F \, \colon \, \omega^\omega \rightarrow \omega^{\omega_1}$ is total. I will use fairly standard notation, i.e. $f,g \in \omega^\omega$, $\alpha, \beta \in \omega_1$ and $k,m,n \in \omega$.
The proof works as follows: Towards a contradiction assume that such a function $F$ exists.
First we will construct a sequence $(a_n)_{n \in \omega} \in \omega^\omega$ and find an ordinal $\beta < \omega_1$ such that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, ,$$ where $\prod_{m \leq n} a_m \times \prod_{m > n} \omega =\{ f \in \omega^\omega \, \colon \, \forall m \leq n \,\, f(m) < a_m \}$. The idea behind this is that we can require a bound for finitely many values of an input function $f$ and still make $F(f)(\beta)$ arbitrarily large.
In the second step we construct a sequence $(b_n)_{n \in \omega} \in \omega^\omega$ such that $(b_n)_{n \in \omega} \geq (a_n)_{n \in \omega}$, and a sequence $(f_n)_{n \in \omega}$ such that $$\forall n \in \omega \, \colon \, f_n \in \omega^\omega \, \land \,(b_m)_{m \in \omega} \geq f_n \, \land \, F(f_n)(\beta) \geq n \,.$$ But this leads to a contradiction, since the monotonicity of $F$ implies that $\forall n \in \omega \, \colon \, F((b_m)_{m \in \omega})(\beta) \geq n$.
First step:
We will construct $(a_n)_{n\in \omega}$ by induction. For the base case $n=0$ we claim that $$\exists a_0 \in \omega \,\exists \alpha_0 \in \omega_1 \,  \forall A \in [\omega_1 \setminus\alpha_0]^{\leq \aleph_0} \, \colon \, F[\prod_{m = 0} a_0 \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A\, .$$ This means that already $F\restriction  \prod_{m = 0} a_0 \times \prod_{m>0} \omega$ dominates $\omega^A$ for every countable $A \subseteq \omega_1$ above $\alpha_0$.
Towards a contradiction assume that the claim is wrong. Therefore, we can construct a sequence $(A_n)_{n \in \omega}$ such that $A_n \in [\omega_1]^{\leq \aleph_0}$ and $\sup A_n < \min A_{n+1}$, and a sequence of functions $(f_n)_{n \in \omega}$ such that $f_n \in \omega^{A_n}$ and $F\restriction  \prod_{m = 0} n \times \prod_{m>0} \omega$ does not dominate $f_n$. But if we define $B:= \bigcup_{n \in \omega} A_n \in [\omega_1]^{\leq \aleph_0}$ and $f:= \bigcup_{n \in \omega} f_n \in \omega^B$, we reach a contradiction, since noting that $\omega^\omega = \bigcup_{n \in \omega} \prod_{m = 0} n \times \prod_{m>0} \omega$, there cannot exist a $g \in \omega^\omega$ such that $F(g) \restriction B \geq f$.
Assume inductively that $a_0,..,a_n$ and increasing $\alpha_0,...,\alpha_n$ have already been defined, such that $\forall A \in [\omega_1 \setminus\alpha_n]^{\leq \aleph_0} \, \colon \, F[\prod_{m \leq n} a_m \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A$. With a similar argument as before, and again noting that $$\prod_{m \leq n} a_m \times \prod_{m>n} \omega = \bigcup_{k \in \omega} \prod_{m \leq n} a_m \times \prod_{m=n+1} k \times \prod_{m>n+1} \omega$$ we can show that $$\exists a_{n+1} \in \omega \, \exists \alpha_{n+1} \in \omega_1 \setminus \alpha_n \, \forall A \in [\omega_1 \setminus\alpha_{n+1}]^{\leq \aleph_0} \, \colon$$ $$F[\prod_{m \leq n} a_m \times \prod_{m=n+1} a_{n+1} \times \prod_{m>n+1} \omega] \,\text{is cofinal in} \, \omega^A \, .$$ So we can find $a_{n+1}$ and $\alpha_{n+1}$ as required.
Now let $\beta > \sup_{n \in \omega} \alpha_n$. It follows from our construction of the $(a_n)_{n \in \omega}$ that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, .$$
Second step:
Again, we will construct $(b_n)_{n \in \omega}$ and $(f_n)_{n \in \omega}$ by induction. The base case $n=0$ is quite easy: Set $b_0=a_0$ and pick any $f_0 \in \prod_{m = 0} b_0 \times \prod_{m>0} \omega$. Then $F(f_0)(\beta)\geq 0$ trivially holds.
Assume inductively that $b_0,...,b_n$ and $f_0,...,f_n$ have already been constructed such that $$\forall m \leq n \, \colon \, a_m \leq b_m \, \land \, f_m \in \prod_{m \leq n} b_m \times \prod_{m>n} \omega \land \, F(f_m)(\beta) \geq m \, .$$
By using what we have shown in the first step, we can now find $f_{n+1} \in \prod_{m \leq n} b_m \times \prod_{m>n} \omega$ such that $F(f_{n+1})(\beta) \geq n+1$. We set $b_{n+1}:= \max( \max_{m \leq n+1} f_m(n+1), a_{n+1})+1$. It follows that $$\forall m \leq n+1 \, \colon \, a_m \leq b_m \, \land \, f_m \in \prod_{m \leq n+1} b_m \times \prod_{m>n+1} \omega \land \, F(f_m)(\beta) \geq m \, .$$
But now we have reached a contradiction, since $\forall n \in \omega \, \colon \, (b_m)_{m \in \omega} \geq f_n$, and therfore the monotonicity (and totality) of $F$ implies that $\forall n \in \omega \, \colon \, F((b_m)_{m \in \omega})(\beta) \geq n$.
