There is no assurance that T(c)<=c $T(c) \leq c$ from T $T$ being a "nonincreasing function". It is quite possible for T(k)>k $T(k) \gt k$ to hold for some cardinals.
It is possible to show that if |phi(c))| $|\phi(c))|$ so is |phi(T(c))| $|\phi(T(c))|$ and also that if |phi(c)| $|\phi(c)|$ is finite and T^{-1}(c) $T^{-1}(c)$ exists (as it would if T(c)>c$T(c) \gt c$) then so is |phi(T^{-1}(c))|. $|\phi(T^{-1}(c))|$. So T(c) < c $T(c) \lt c$ is impossible because c $c$ is the smallest cardinal of the kind indicated, and T(c) > c $T(c) \gt c$ is impossible NOT because "T $T$ is a nonincreasing function" but because we would then have T(c)>c>T^{-1}(c) $T(c) \gt c \gt T^{-1}(c)$ and T^{-1}(c) $T^{-1}(c)$ would be a cardinal of the given sort smaller than c.$c$.
THUS we have c=T(c) $c=T(c)$ and so |phi(c)| $|\phi(c)| = |phi(T(c))|. \phi(T(c))|$. But also we can show that |phi(T(c))| $|\phi(T(c))| = T|phi(c)|+ T(|\phi(c)|)+$ (1 or 2) [NOT |phi(c)|+ $|\phi(c)|+$ (1 or 2)]. Standard natural numbers are fixed by the T $T$ operation, but general natural numbers do not have to be. We can show that |phi(c)| = T|phi(c)| mod 3$|\phi(c)| \equiv T(|\phi(c)|) \pmod 3$, and this is enough to get the contradiction.
Unfortunately, the argument just isnt isn't intuitive.
