The purpose of the diode is quench the stored energy in the switch with an inductive load. Your transistor is the switch which could in theory switch the output better than the input to the boost converter. Presuming that the no load current is ok when the fan is off and boostconverter still running.

• it is wise to move the switch to the output and switch the fan -ve with the diode you have placed there. The fan load appears like a damped resistive load with a capacitance in parallel as it commutates at some freq. due to internal caps. (.i.e. the current rises linearly with voltage)
• yet accelerating the voltage changes will draw more current until steady state.

The purpose of the diode is quench the stored energy in an switched off inductive load. 

Your transistor is the switch so it needs the diode, ... which could in practice will switch the output better than the input to the boost converter. Presuming that the no load current is ok when the fan is off and boostconverter still running.

• it is wise to move the switch to the output and switch the fan -ve with the diode you have placed there. The fan load appears like a damped resistive load with a capacitance in parallel as it commutates at some freq. due to internal caps. (.i.e. the current rises linearly with voltage)
• yet accelerating the voltage changes will draw more current until steady state.
• The boost converter will have more surge current than the fan . This is why it is better to switch the output and easier on the low side of the load.