Operation of a Monostable circuit

In its stable condition (T0 - Time 0)), Q1 is cut off and Q2 is conducting.
The input trigger (positive pulse at T1) is applied to the collector of Q1 and coupled by C1 to the base of Q2 causing Q2 to be cut off (Since these are PNP transistors, a negative signal is needed for the transistor to operate).

The collector voltage of Q2 then goes −VCC (This is because R5 produces a voltage drop and the juntion between R5 and the collector of Q2 is the source voltage -Vcc).

The more negative voltage at the collector of Q2 forward biases Q1 through R4 (this means that Q1 has a more negative signal applied to its base viaR4).

With the forward bias, Q1 conducts (from its off state), and the collector voltage of Q1 goes to about 0 volts (remember, for a transistor to operate, the base voltage is forward biased whilst the collect leg is Reverse biased - Now this means that, for a PNP transistor - Common Emitter, when Q1 is off, the Collector is -Vcc (because the resistor has the -Vcc voltage drop and this is taken with the junction of the collector leg, and the Transistor has a 0 Voltage Drop) - When Q1 turns on, the current at the collector increase, which causes a voltage drop across the transistor, which in turn reduces the current (and hence voltage) at the R3/Collector junction (from say -5v to 0v).

C1 now discharges and keeps Q2 cut off. Q2 remains cut off until C1 discharges enough to allow Q2 to conduct again (T2). When Q2 conducts again, its collector voltage goes toward 0 volts (because of the voltage drop across Q2) and Q1 is cut off. The circuit returns to its quiescent state and has completed a cycle. The circuit which in turn remains in this stable state until the next trigger arrives (T3).