Small-signal operation of the BJT differential pair
You can think of this as an assumption based on symmetry and linearity. If the differential input voltage is very small (&lt;&lt; 2Vt) then the input resistance of the two transistors will remain constant and the small differential voltage will divide equally. 

The \$\alpha\$ factor is close to one for real transistors, so Ic ~= Ie. 

They're really just saying that the collector currents are about I/2 in each case, and they increase/decrease by the transconductance gm multiplied by half the input voltage, and the collector voltages decrease/increase by the gm*Vid/2 times the collector load resistances. 

Not that they have not separated the large signal bias from the small signal behavior in this example.