- #1
laminatedevildoll
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If f X---> Y is a function and if there is exactly one function g:Y---> X so that
f o g = id_y, the f is a bijection and g=f^-1. Do I need to use the axiom of choice to prove this theorem?
f o g = id_y, the f is a bijection and g=f^-1. Do I need to use the axiom of choice to prove this theorem?