Does there exist a section of $\mathcal{P}(\kappa)\to\mathcal{P}(\kappa)/(\text{fin})$ that is "nearly Boolean"?

Question

The following might be a somewhat esoteric question:

Does there exist an infinite cardinal $\kappa$ and a section $f$ of the quotient map $\pi:\mathcal{P}(\kappa)\to\mathcal{P}(\kappa)/(\text{fin})$ (i.e., $f:\mathcal{P}(\kappa)/(\text{fin})\to\mathcal{P}(\kappa)$ and $\pi\circ f=\text{Id}_{\mathcal{P}(\kappa)/(\text{fin})}$) s.t.,

1. $f$ is monotone;
2. If $x_1,\cdots,x_n\in\mathcal{P}(\kappa)/(\text{fin})$ are such that $\bigvee_{i=1}^nx_i=[\kappa]$, then $\bigcup_{i=1}^nf(x_i)=\kappa$;
3. For any ultrafilter $\mathcal{U}$ on $\mathcal{P}(\kappa)/(\text{fin})$, $\bigcap_{x\in\mathcal{U}}f(x)=\varnothing$.

If this is not possible, what if $\mathcal{P}(\kappa)/(\text{fin})$ is replaced by a further (Boolean algebra) quotient? (That will already be enough for my purposes.)

This is related to my attempt at a topology question. One of my approaches at constructing a counterexample for that question requires the existence of a section $f$ as above. (The precise connection is not really relevant to the present question.) It is not immediately clear to me that such an $f$ should not exist - I'm aware that if $f$ is required to be a Boolean algebra homomorphism, this is probably impossible (at least when $\kappa=\omega$), but conditions $1$ and $2$ are a priori strictly weaker than asking for a Boolean algebra homomorphism.

Answer 1

No, this is not possible. In fact, you don't even need conditions 2 and 3 from your list, because it is already impossible to have a monotone section of the quotient map $\pi: \mathcal P(\kappa) \rightarrow \mathcal P(\kappa)/(\mathrm{fin})$.

To see this, fix a countably infinite $A_0 \subseteq \kappa$. Using a length-$\omega_1$ recursion, you can build a sequence $\langle A_\alpha :\, \alpha < \omega_1 \rangle$ of subsets of $A_0$ such that for all $\alpha < \beta < \omega_1$, $A_\alpha \setminus A_\beta$ is infinite and $A_\beta \setminus A_\alpha$ is finite. (Such a thing is sometimes called an $\omega_1$-tower.) If $f$ were a monotone section map, then $\langle f(A_\alpha) :\, \alpha < \omega_1 \rangle$ would be an $\omega_1$-sequence of countable sets such that if $\alpha < \beta$ then $f(A_\alpha)$ is a proper subset of $f(A_\beta)$. And this is clearly impossible.