 Hello and welcome to the session. Let us cause the following problem today. Let a and b be set, show that f is a function such that from a cross b to b cross a such that f of a comma b is equal to b comma a is a bijective function. Now let us write the solution. Now let us check for one one. Let a one comma b one comma a two comma b two belongs to a cross b such that f of a one comma b one is equal to f of a two comma b two which implies b one comma a one is equal to b two comma a two which implies b one is equal to b two and a one is equal to a two which implies a one comma b one is equal to a two comma b two thus f of a one comma b one is equal to f of a two comma b two which implies a one comma b one is equal to a two comma b two for all a one comma b one and a two comma b two belongs to a cross b thus f is one one. Now let us check for on two. Let b comma a be an arbitrary element of b cross a then b belongs to b and a belongs to a which implies a comma b belongs to a cross b thus for all b comma a belongs to b cross a there exists a comma b belongs to a cross b such that f of a comma b is equal to b comma a so f is a function such that from a to b to b cross a is an on two function thus f is an on two function therefore f is a bijective function since it is one one and on two both. I hope you understood this problem. Bye and have a nice day.