Looks like a free wheeling diode for an inductive load connected to the “output” terminals. The GND2 symbol makes no sense as shown, adding to the confusion.
Is the main reason for that diode to protect the mosfet for inductive load?
Yes, this diode protects the MOSFET from self-induction emf when the inductive load is disconnected is switched off by the MOSFET. If the load is active, it is not needed.
Thanks. Then, i understand 
No, not disconnected, but when it is switched off by the FET.
If the inductive load is disconnected by parting the connector, it may cause an arc, but it will not apply a reverse voltage to the circuit.
"Lost in translation" by Google Translator.
Of course, "the inductive load is disconnected", should be read as "the inductive load is switched off by the MOSFET".
Thank you.
Actually arcs are very destructive, sometimes generating high levels of RF voltage, so it may apply a whole mess of alternating voltages, so both forward and reverse over-voltage can happen - a TVS diode or MOV might be needed to increase the protection against an arc - a freewheel diode isn't enough.
But this is not going to be relevant here as this can only happen while the FET is switched on and "locking" the voltage at the output terminals to the supply voltage. 
Just to note that the MOSFET already has a body diode that'll protect it when switching inductive loads. You should probably remove the GND2 symbol (output pin2) as this is actually the switched output (LOAD) signal (see reply#2). Also, output pin 1 is actually +DC power.
Sadly this is not true.
Only in an H-bridge the other MOSFET may be used as a freewheeling diode.
Gladly, it really is true, that the body diode protects it (the MOSFET). I didn't say to remove the IN4007 which is still needed in this low-side switching application.
How does the body diode protect the MOSFET? By avalanche breakdown? Or you mean different schematic than OP?
also known as a commutating diode. people get weird about this, it has multiple names and people believe the name they heard first is THE name.
I'm puzzled. How does the body diode protect the MOSFET?
Reverse spikes (connecting or disconnecting the inductive load or power or operating with defective external flyback diode).
I am not claiming anything, just a quote.
The problem can arise when an inductive load is used at the drain side of the device, and during the MOSFET switch OFF periods the inductor's reverse EMF passing through the MOSFET body diode becomes too high, causing a sudden rise in the MOSFET's junction temperatures, and its breakdown.
I still don't understand. An inductive load doesn't create a reverse EMF over the switching transistor, it creates the reverse EMF over the inductor itself.
At the transistor, this looks like a high forward EMF. Assuming a positive supply and a low-side switch, the inductor creates a very high positive voltage (higher than Vcc) at the drain/collector when the transistor is switched off. With a high-side switch, the inductor creates a high negative voltage (lower than GND) at the emitter/source. In both cases this reverse biases the internal diode.
That is why, when a switching transistor needs protection, the protection diode is never connected across the transistor, it is connected across the load. A diode across the transistor does nothing to protect it. This is very well known.
What you say is correct.
It is known that a flyback diode connected in parallel with an inductive load reduces the self-induction emf to the level of the forward voltage drop across the diode and somewhat slows down the current decrease. This is a well-known practice that has been in use for decades. What is there to argue about is unclear.
