The base to emitter voltage is pretty stable at 0.6V, you are not adding any feedback with the resistor.
You actually built a emitter follower or common collector. The main property of this circuit is that the emitter is pretty much at the same voltage as the base voltage.
If you are hinting at the non ideal behavior of the transistor, even the slight variation in BE-voltage is useless as an emitter follower has a gain less than 1, it actually attenuates slightly. So instead of adding feedback, you are actually just adding an extra load to the input.
By using two transistors for a latch, you actually add gain to the circuit and you add positive feedback.
The circuit in the image doesn't use the emitter follower, it uses two common emitters, which actually does have substantial voltage and curent gain.
Assume the input is low and the relay is not activated. Because there is no current flowing through the relay, R3 is pretty much grounded. When you start pulling the input trigger up slowly, at about 0.7V input voltage the leftmost transistor will start to conduct, causing T2 base to be pulled down. T2 will start to conduct and causes an ever larger current through the relay. Because of the relatively high current through the relay, R3 will be pulled up to a higher voltage and cause the input to be pulled up even further as a result. Because of the resulting higher input voltage, T1 will conduct even more, pulling T2-base even harder down, causing more current through the relay, eventually swinging the relay voltage all the way up to the power supply rail.
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