Path from correspondence between fermionic oscillator and bosonic oscillator to free bosons from interacting fermions This question is open to edits and corrections. I'm available to make corrections as needed.
I am attempting to understand fermion boson correspondence and bosonization.
What is the correspondence between the spectrum of the fermionic oscillator and the bosonic oscillator? I think it has a definite answer if it exists of the form  $E_f = A * E_b$, and perhaps a correspondence exists too between the states.

So $E_f = (n - 1/2)$
      = { -1/2,1/2,3/2,5/2  . . .  }  but should one terminate at 1/2

and  $E_b = (n + 1/2)$
      = {     1/2,3/2,5/2 . . . . .}

but it seems to be the case that n can only be 0 or 1  for $E_f$ but for $E_b$ n goes from 0 to $\infty$
Also, the "number states" can be found for both cases and compared.
On the level described here, it seems to me that there isn't much happening.
What about at the level of the ladder operators?

If a field can be written in the form  $\psi^{\pm} \propto  c^{\pm}$ and
a hamiltonian built from $\psi^{\pm}$
then perhaps bosonization can be done.
I feel like these ingredients are enough to put together coherent sentences about how to bosonize a theory. I just have not figured it yet.

I don't know much about the fermionic oscillator except what I think Pauli exclusion means  $b^+b^+ | 0\rangle = 0$ and $b | 0 \rangle = 0$
How does one rewrite the fermionic operators as bosonic ones?
I've come across the terms fermion boson correspondence and bosonization. I'm not going to attempt to ask for references on the subject because I'm likely to not truly understand them. What struck me about these sorts of things is that it seems possible to turn a theory of interacting fermions into a theory of free bosons.
If the simplest interaction is then cooked up in 1d for two fermions, derived in some way from whatever scheme I am attempting to prepare here, how can the theory be written as free bosons?
I hope we can do this with basic ladder operators and no path integrals, please.
 A: This is what happens in the description of superconductivity: the Cooper pairs, two electrons in interaction gives a boson of spin 0, so a phase transformation is needed to have two fermions (e-) of spin 1/2 .

If the simplest interaction is then cooked up in 1d for two fermions, derived in some way from whatever scheme I am attempting to prepare here, how can the theory be written as free bosons.

In this link in French (2.1.3), he starts from the following remark:

Therefore, the Cooper pair is a scalar particle, and we can
using the Lagrangian of a scalar field coupled to an electromagnetic field to break
the symmetry...

and he comes to the conclusion :

In addition, note again that, since $\phi$ represents a particle composed of two electrons, the coefficient a and therefore the parameter $\mu^{2}$ can be interpreted as
a coupling constant between the two electrons, or between an electron and a phonon. In
In general, the coupling constant between particles depends on the nature of the interaction
prevailing between the particles in question. Thus, $\mu^{2}$ is no longer a parameter whose value
must be given by hand, but a quantity whose value can be calculated from
of the interaction between the electrons forming the Cooper pair. Gold, for example,
is a good electrical conductor at room temperature, which means that the constant of
coupling between electrons and phonons is weak at room temperature. Even at low
temperature, the coupling constant for gold remains low. Therefore, gold is not a
superconducting material, since a strong coupling between electrons and phonons
is necessary for the formation of Cooper pairs, which would then break the U(1) symmetry
and thus trigger superconductivity. The quantity $\mu^{2}$ therefore depends on the dynamics
between the electrons, and this is why we speak of a dynamic symmetry breaking.
(google tr...)

If I didn't understand the question, there are distorted (supersymmetric) quantum groups.
