Cross-entropy via softmax: Mathematical derivation I am trying to understand the derivation of cross entropy loss in the context of softmax. However, some steps are still not clear to me. Hence, I would appreciate if someone could please explain.
y = one hot encoding (representing true label) = selector
$y ̂= $value predicted by NN and the softmax.
Through the NN:
$X\xrightarrow{\smash{NN, θ}} Z \xrightarrow{\smash{Soft-max}} \hat{y}$
To derive the loss function, we start out from the likelihood function that a given set of parameters θ of the model can result in prediction of the correct class of each input sample. The maximization of this likelihood can be written as:     $$ arg⁡ max _θ L(θ|y ̂,Z) $$
The likelihood $ L(θ|y ̂,z) $ an be rewritten as the joint probability of generating t and z given the parameters θ, which can be written as a conditional distribution:
$$ L(θ│\hat{y} ,Z)=P(\hat{y} , Z│θ)=P(y ̂│Z,θ)P(Z│θ) $$
Since we are not interested in the probability of z we can reduce this (assumption 1):
$$ L(θ│y ̂,Z)∝P(y ̂│Z,θ) $$
Which can be written as $P(y ̂|z) $ for fixed θ (assumption 2).
Since each yc is dependent on the full z and only 1 class can be activated in the t we can write
$$ P(y ̂│Z)= ∏_{(c=1)}^C(P(y ̂_c |Z))^{y_c} =∏_{(c=1)}^C((SoftMax(Z))^{y_c }=∏_{(c=1)}^C(y ̂_c )^{y_c} $$
Assumptions 1 and 2 are not clear to me. Some explanation would be appreciated.
Hints:

$P(Z|\theta)$ is independent of $\hat{y}$ so it can be left out as merely a multiplicative factor as we are interested in $\hat{y}$.
$\theta$ is a parameter and for each model is indeed fixed, it is not a variable