[ExI] Electronic Circuit Question (Emitter Follower)

[Lee Corbin]

I have a question for people who know about transistor
circuits. On page 56 of Horowitz and Hill's classic "The Art
of Electronics", there is a nice description of the emitter
follower circuit.  The following is used as an example:

The bottom of the diagram is at -10 volts and the top is at
+10 volts (i.e. a 20volt supply somewhere). Just above the
-10 volts is a 1K resistor, and above that the emitter of an
NPN transistor.  There is no resistor between the collector
and the +10 volts.  The experiment is to let the base voltage
(input) vary between +10 and -10.  The output is taken
(hence "emitter-follower") at the top of the 1K resistor.

Because the base-emitter voltage is always around .6 volts,
the output naturally follows the input, but at .6 volts less.

But the book says that when the input voltage drops down
to -4.4 volts, the base-emitter junction gets back-biased,
(and the transistor turns off?).  I don't understand why the
voltage on the base cannot keep going down, say to -6V,
with the output voltage continuing to keep in step, say at
-6.6.  Even at -6 volts, there seems to me to be plenty
of leeway between that and the -10V source below it.

Here is their explanation:

   "The output can swing to within a transistor saturation
   voltage drop of VCC (about +9.9v) but it cannot go
   more negative than -5 volts. That is because on the
   extreme negative swing the transistor can do no more
   than turn off, which it does at -4.4 volts input (-5V
   output).  Further netgative swing at the input results in 
   back-biasing of the base-emitter juntion, but no further
   change in output."

I still don't see why the base could not be at, say, -6v
and the output .6 lower.  Why should the base-emitter
junction be back-biased when there is still a big voltage
difference between the base and the -10 volts at bottom?

Thanks,
Lee