Work done on a container, resulting in pressure change Why does pumping gas into a rigid container (causing the pressure and temperature to increase), not mean that work has been done on the container?

Work is given by the formula :  W = -P ΔV

If I go off the formula alone, ΔV=0 (as the container is rigid) and hence W=0

But if I ignore the formula and take the definition of work as "the energy it takes to move an object against a force", wouldn't pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container e.g. like pumping a bike tyre, the gas is not going to go into the container unless I pump it in?



For reference: 

Question from textbook: A rigid container of constant volume is used to store
compressed gas. When gas is pumped into the container,
the pressure of the gas inside the container is increased and
the temperature of the container also increases. Which
statement is true of the work done on the container? 

(a) The work is equal to the increase in the pressure
inside the container.

(b) The work is equal to the increase in the temperature inside the container.

(c) The work is equal to the sum of the pressure and temperature increases.

(d) There is no work done on the container. 

Answer from textbook:  (d) the container does not expand, so as there is no
change in volume of the container, no work is done on the
container.
Yes, the explanation 


  the container does not expand, so as there is no change in volume of the container, no work is done on the container.  


is right as it takes into account that the container is rigid, so its volume can not be changed. Thus, there is no work done by the gas on the container.

Also, if you say


  " pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container "


Yes, there is work done on the gas, but it is by the piston(or the pump).

You should see how the laws of Thermodynamics make clear about the boundary between system and its surroundings, so work has to be done by "something" on "something".
Also, In the question, it is clearly asked "Which statement is true of the work done on the container?" So, as the container's volume is unaltered, there is no work on it.