I'm asking my question for a general symmetric monoidal $\infty$-category $ V$ and a general indexing simplicial set $I$, but my specific interest is for $ V = Sp$ with the usual smash product and $I = \Delta^{op}$ (more precisely its nerve), so if the answer is different for these choices, I'd be glad to hear it too. I think the answer is no, but I'm not entirely sure and can't come up with a specific counterexample.

I have two functors $A,B : I\to Alg_{E_\infty}( V)$ where $Alg_{E_\infty}$ denotes the $\infty$-category of commutative algebras in $V$, and if I let $U : Alg_{E_\infty}( V)\to V$ denote the forgetful functor, I have a map $f: UA \to UB$, with, for each $i\in I$, a given lift $\tilde f_i : A_i\to B_i$ of $f_i : UA_i \to UB_i$.

Can I then (up to equivalence in $V^I$) lift the whole $f$ to a map in $Alg_{E_\infty}(V)$ ?

In other words, let $Alg_{E_\infty}(V)^I \to V^I \times_{\prod_{i\in I_0} V} (\prod_{i\in I_0}Alg_{E_\infty}(V))$ be the obvious functor, is is surjective on $\pi_0 Map$ ?

Again, if this helps, I'm willing to assume $V=Sp, I=\Delta^{op}$ (but I think this case won't be too different from the general case)

I'm also interested (although somewhat less) in the same question where you remove "symmetric", and replace $E_\infty$ with $E_1$, but again, I think the answer should be pretty much the same (probably no).

Of course the best of cases would be in the given situation, not even being able to find an $E_1$-lift.

I'm asking my question for a general symmetric monoidal $\infty$-category $ V$ and a general indexing simplicial set $I$, but my specific interest is for $ V = Sp$ with the usual smash product and $I = \Delta^{op}$ (more precisely its nerve), so if the answer is different for these choices, I'd be glad to hear it too. I think the answer is no, but I'm not entirely sure and can't come up with a specific counterexample.

I have two functors $A,B : I\to Alg_{E_\infty}( V)$ where $Alg_{E_\infty}$ denotes the $\infty$-category of commutative algebras in $V$, and if I let $U : Alg_{E_\infty}( V)\to V$ denote the forgetful functor, I have a map $f: UA \to UB$, with, for each $i\in I$, a given lift $\tilde f_i : A_i\to B_i$ of $f_i : UA_i \to UB_i$.

Can I then (up to equivalence in $V^I$) lift the whole $f$ to a map in $Alg_{E_\infty}(V)$ ?

In other words, let $Alg_{E_\infty}(V)^I \to V^I \times_{\prod_{i\in I_0} V} (\prod_{i\in I_0}Alg_{E_\infty}(V))$ be the obvious functor, is it surjective on $\pi_0 Map$ ?

Again, if this helps, I'm willing to assume $V=Sp, I=\Delta^{op}$ (but I think this case won't be too different from the general case)

I'm also interested (although somewhat less) in the same question where you remove "symmetric", and replace $E_\infty$ with $E_1$, but again, I think the answer should be pretty much the same (probably no).

Of course the best of cases would be in the given situation, not even being able to find an $E_1$-lift.