Boost converter, how does diode allow current from inductor to flow towards capacitor?
An inductor works on the following fundamental principle: -

\$V = L\frac{di}{dt}\$ 

Where \$\frac{di}{dt}\$ is the rate of change of current in the inductor

If you apply a voltage across an inductor (by grounding one end for instance), current will climb at the rate V/L amps per second and, while doing so, energy will be stored in the magnetic field. When the inductor is disconnected from that ground connection, the stored energy pushes a current out of the inductor in the same direction.

That current tries to maintain its value but it can't and so current starts to fall. This means that the rate of change of current (\$\frac{di}{dt}\$) is negative. 

This generates a negative voltage across the inductor terminals \$V = L\frac{-di}{dt}\$.

The input side of the inductor is "tied" to the incoming supply voltage hence, the switched side of the inductor (previously at ground) generates a voltage that is greater than the incoming voltage - this is the voltage reversal seen across the inductor i.e. this is a negative voltage compared to when the inductor was grounded by the transistor.

This "greater voltage" rapidly rises (in order to push current out) and when this rapidly rising voltage equals the voltage across the output capacitor (plus one diode drop) it finds a "load" to dump the current into. From this point on the output voltage acquires a level suitable to carry on pushing current into the capacitor until all the magnetic energy previously stored is depleted.