Inverting Op-Amp Schmitt trigger threshold
Am I right with my objection?  No , and here's why.



Consider the three circuits. 

(1) Without hysteresis we have a simple potential divider and a single threshold.

(2) Adding R3 to the output means that when the output is high (assume = V) then R3 is effectively in parallel with R1  (R1//R3). This gives the high threshold. When the output is low (assume = 0) then R3 is effectively in parallel with R2 (R3//R2). This gives the low threshold value.

(3) Removing (omitting) R1 still leaves us with a potential divider formed from R3 and R2. When the output is low (assume = 0) this gives us 0 x R2/(R2+R3) which is 0. The high threshold is given by V x R2/(R2+R3).

This is exactly what is implied in the book.

Omitting R2 instead of R1 would mean that the + input could never be pulled down to ground but would either be +V or V x R3/(R1+R3)

Note that the assumption is for an ideal op amp. (i.e. V out switches between rail voltages)