I am an EE , so I'll throw in my two cents.
Warning: lesson in circuits ahead! Also, note that I'm tired and may have made some mistakes, so I'll state the disclaimer than if you rely on this advice, you may electrocute yourself.

The voltage divider formula is correct on its own, but slightly incomplete. It works perfectly when the device that uses the "tap" has infinite input resistance (impedance), but that's not necessarily a good assumption. Let's call the device's internal resistance RL (L for Load)

I'll use some toy numbers which might be reasonably close to your setup (but definitely aren't exact, since I'm pulling them out of my head)

For example - let's say R1=100 Ohm, R2 = 900 Ohm, and RL is something infinite (or maybe just huge, relative to R2 -- here, assume it's 1 MOhm = 1000000 Ohms):


+ --------R1-----o----------- +
| |
V_in R2 RL VL
| |
- ------------------o----------- -

(The vertical bars here were supposed to connect R2 and RL to the circuit in parallel)

The two resistors R2 and RL are in parallel, so they have an equivalent resistance defined as 1/R_eq = 1/R2 + 1/RL.

The equivalent circuit has R1 and R_eq in series, tapping VL off of R_eq.

If RL is infinite, then 1/RL = 0, so R_eq = R2. In the finite (but big) case:
1/R_eq = 1/900 + 1/1000000, so R_eq is basically still 900 Ohms (since RL is so much bigger than R2, then 1/RL is much smaller than 1/R2.)

Problem: In the real world (and on your camera), the device's impedance is probably something much smaller, so we need to take it into account. Let's say it draws 100 mA constantly while charging (something that could be checked) -- this means it has a load resistance of 4.5 V / 100 mA = 45 Ohms.
This isn't very big, but let's pick R1=1 Ohm and R2=9 Ohms to help our case (so that RL > R2).

In this case, R_eq = 9*45/(9+45) = 7.5 Ohms, so we get VL = 7.5Ohm/(7.5Ohm+1Ohm) * 5 V = 4.4 V. Pretty close to what we wanted - however, this leads to another problem.

Another problem: Since we had to pick R2 much smaller than RL in order for the divider to work, it now draws a current of VL / R2 = 4.4V / 9 Ohms = 488mA. That means that the power supply needs to be able to provide 588 mA of current to its load, and most of it is being wasted as heat! Power = V*I = 4.4V*0.488A > 2 Watts! (This also means you need a R2 resistor rated for at least this much power dissipation)

The moral of the story -- yes, a simple voltage divider may work, but even if it does, it will waste a lot of power as heat. This analysis also relies on the fact that the 5V provided as the input is constant -- it is possible than if the power supply isn't regulated, its output voltage may change based on the load, which messes with all the previous analysis.

This post is already getting way too long -- if you want to discuss it further, feel free to send me a PM.