Is $Z=e^{-beta F}$ as general as $Z=sum_{Ei}Omega(E_i)e^{-beta E_i}$?

I’m having some trouble understanding the formula: $Z=e^{-beta F}$ which is sometimes used to define the partition function.

From what I understand, this $Z$, is the partition function but taken at the thermodynamic limit since:

$$Z=sum_{E_i}Omega(E_i)e^{-beta E_i}=sum_{E_i}e^{S_m(E_i)/k_b}e^{-beta E_i}=sum_{E_i}e^{beta(S_m(E_i)T- E_i)}=sum_{E_i}e^{-beta F(E_i)}$$

And then, by taking the thermodynamic limit; we would find that $F$ is minimal at $E^*$ (which is approximatively equal to $bar{E}$) and because $F$ is (inversely) peaked at $E^*$ we could write:

$$Z=sum_{E_i}e^{-beta F(E_i)}simeq e^{-beta F(bar{E})}$$

And then, saying that $Z=e^{-beta F}$ would be an approximation a bit like writing: $Z=Omega(bar{E})e^{-betabar{E}}$.

Is this right? Or am I missing something?

note:

I’am not even sure I can write $E_i-TS_m(E_i)=F(E_i)$ since $S_m$ is the microcanonical entropy and not the thermodynamic one nor the canonical one (I guess?)…

Is it right too, to write $F$ as a function of the energy, in this case? Since it is supposed to be a function of $V$ and $T$?

We could also define $F$ from: $F=bar{E}-TS$ and then, from $S=k_bfrac{partial Tln(Z)}{partial T}$ and $bar{E}=- frac{partial ln Z}{partial beta}$ but we are still using a thermodynamic relation and not an "exact" one?