The constraint that the n outputs must sum to $1$ means that only $n1$  parameters are necessary; the probability of the $n^{th}$ value may be  obtained by subtracting the first $n1$ probabilities from $1$. We can  thus impose a requirement that one element of $z$ be fixed. For example,  we can require that $z_n = 0$. Indeed, this is exactly what the sigmoid  unit does. What does it mean by impose a requirement that one element of $z$ be fixed. I understand you can get the $n^{th}$ one if you know n-1 but fixed?Defining $P (y = 1 | x ) = \sigma(z )$ is equivalent to defining $P(y = 1 | x) = softmax(z)_1$ with a two-dimensional $z$ and $z_1 = 0$. Both  the $n  1$ argument and the $n$ argument approaches to the softmax can  describe the same set of probability distributions, but have different  learning dynamics.How is it equivalent to having $z_1=0$?
Softmax function forms
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