Q:

Parallel a diode to solid state relay load is a short circuit?

I'm trying to use a solid state relay control a 12v dc air pump, so I was reading the "Solid State Relays Common Precautions"(like below).
http://omronfs.omron.com/en_US/ecb/products/pdf/precautions_ssr.pdf
When I read the paragraph shown in the screen, I'm wondering if I parallel a diode to the load, I'm I making a short circuit since there won't be much resistance in the dioade so the current will just bypass the load and go through the dioade?

A:

You need to connect the diode the correct way (reverse biased, as shown), otherwise it will indeed conduct when the SSR turns on and likely destroy the SSR and quite possibly the diode. 
When connected as shown, it is reverse biased until the SSR turns off, upon which the inductive load will cause the diode to become briefly forward biased until the energy in the magnetic field is absorbed by the coil resistance and diode forward drop. Otherwise the voltage across the load would increase to potentially damaging (to the SSR) levels. 
Consider the below schematic- the switches represent the SSR and the stuff in the boxes represents your pump motor. 

simulate this circuit – Schematic created using CircuitLab
The switches open at t=0 and current is 1A through each circuit at that time. As you can see in the below simulation, the voltage across the switches increases to a bit over 12V and then drops back to 12V when the diode stops conducting about 2.5ms later. 

If I remove D2, the difference is dramatic: 

The voltage across SW1 spikes to about 900V, before dropping back. 

