thermodynamics - Why is the the differential of Helmoltz free energy dT dependent?

My question is simple : we use Hemloltz free energy "A" to study equilibrium of system under transformation at T and N constants.

We have A=E-T.S, but dA=dE-TdS-SdT

Why is the differential dT dependent as we construct this function to study equilibrium for system where the temperature is imposed ?

[edit] : More detailed question :

In fact what I understood is that for any system where the external work is only done by pressure forces we have :

$$ dU+PdV-TdS < 0$$ (it comes from first principle and using the fact that $dS=\frac{\delta Q}{T} + \delta S^c$ with $ \delta S^c > 0$).

And we define a thermodynamic potential $ \Phi $ as $d \Phi = dU+PdV-TdS$. The equilibrium will then be done if $ d \Phi =0$

It is not possible to define a general $ \Phi $, but if we work in (T,V,N) ($dT=dV=dN=0$), we have : $ d \Phi = dU+PdV-TdS = d(U-T S)$, so $ \Phi = A = U-T S$.

Then to construct F we assumed $dT=0$, so why do we consider after that $dT$ is not equal to 0 ? I don't get the logic.