Relation between crossover freqquency of loop gain and closed loop's corner frequency when feedback is not unity
It is not a mistake, the author is correct.


  ..at \$\ f_c\$ (where \$\ T=1\$, with associated \$ 90^{\circ}\$. phase lag), the closed loop gain \$\ G(s)\$ is 3dB down..


Let \$\ T=1\angle90^{\circ}=\cos(90^{\circ})+j\sin(90^{\circ})=j\$

The closed-loop transfer function is:

\$\ G(s) = \frac{1}{K_{FB}} \cdot \frac{T}{1+T}\$ 

Substituting \$\ T=j\$ and solving for the gain:

\$\ |G(s)|= \frac{1}{K_{FB}} \cdot |\frac{j}{1+j}|=\frac{1}{K_{FB}} \cdot |\frac{1}{2}+\frac{1}{2}j|=\frac{1}{K_{FB}} \cdot \sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2}=\frac{1}{K_{FB}} \cdot \frac{1}{\sqrt{2}}\$

\$\ 20 \cdot \log_{10}(\frac{1}{\sqrt{2}})=-3 \$ [dB]

As the author stated; when the open loop gain \$\ T(s)\$ equals one, with associated \$ 90^{\circ}\$ phase lag, the closed loop gain \$\ G(s)\$ is down 3dB, or \$\ \frac{1}{\sqrt{2}}\$