Creating a momentary low pulse with a toggle switch
Solution 1

A simple 1 transistor circuit:



When SW1 is closed it pulls one side of C1 to 0V. The other side of C1 is at 0V so in order to maintain the voltage across C1 it pulls this plate to -V.

D1 (a general purpose diode) prevents this negative going edge exceeding the base-emitter breakdown voltage of T1 for supply voltages greater than about 5V. 

This plate of C1 then charges back up through R2 to about 0V. 

When the switch is closed that side of C1 rises immediately to +V causing the other side of C1  to do the same (can't instantly change the voltage across the capacitor). This side of then C1 discharges through D1 and Vbe in parallel with R2 causing T1 to switch on for a short time. 

Once this plate voltage falls below 1.2V (2 x 0.6V drops) T1 turns off (again) leaving the output voltage at the collector of T1 high. 

Solution 2 

The  555 monostable 

As it stands the standard monostable needs to be tweeked before it will answer this question.



Tweek 1 - is to produce a short negative edge pulse from the switch. (R1,C1,D1 and R2). 

R2 holds the trigger and threshold inputs high. When the switch is closed the negative edge of the pulse is transferred across C1 pulling the voltage at C1, R2 low - triggering the 555. 

Opening the switch will not trigger the 555 and any positive edge pulse produced is clamped by D1 to no more than 0.6V above the supply.

R3 and C2 control the timing of the pulse (standard timing elements- adjust values as required).

C3 is optional but recommended (smoothing of internal divider resistor chain)

Tweek 2 - The output of the 555 is normally LOW with the pulse going high. As you require the opposite (high-low-high) a simple transistor inverter circuit is added.