Pulldown resistor on output pin, how does output still work?
You have two modes.

Input
The input is high-impedance, so there won't flow any current, apart from a small leakage current, which we'll ignore for the moment. Let's say you have a 10 k&Omega; pull-down resistor. Since there doesn't flow any current into or out of the input there won't be any current through the resistor, and then, due to Ohm's Law there won't be any voltage across it either. So if the low end is 0 V, so will be the input. The controller sees it as a low level.

Output
Whether the output is high or low, it's low impedance, like for instance 10 &Omega;. Low won't be a problem: the pull-down already made the level low, and the low impedance of the output only enforces this.  

If the output is high, the internal 10 &Omega; resistor and the 10 k&Omega; pull-down form a resistor divider. The output voltage will then be

\$ V_{out} = \dfrac{10 k\Omega}{10 k\Omega + 10\Omega} 5 V = 4.995 V \$

So the pull-down resistor changes the output voltage only very slightly, thanks to the big difference in resistance.