MATH.6:5.2 - A left inverse and the elements it does not recover

Consider the claim: if f:A→B has g:B→A with g(f(a))=a for every a, then f(g(b))=b for every b.

Choose A={u}, B={0,1}, f(u)=0, and g(0)=g(1)=u. Both maps are total. The premise holds at the only element of A, but f(g(1))=0. The claim fails because the left-inverse condition constrains recovery of source elements while B also contains an element outside the image of f.

Now require A and B to be the same finite set. The premise makes f injective: equality f(a)=f(a') gives a=a' after applying g. An injective self-map of a finite set is surjective. Write any b as f(a); then f(g(b))=f(g(f(a)))=f(a)=b. This supplies a proof for the changed domain.

For the same infinite set N={0,1,2,...}, take f(n)=n+1, g(0)=0, and g(n+1)=n. Then g(f(n))=n for every n, but f(g(0))=1. Every finite self-map search can miss this failure because the finite claim is true. The symbolic construction locates the lost premise: finiteness supplied the step from injection to surjection.

A receiving construction can retain the original left inverse for source recovery, require surjectivity for recovery of every target element, or restrict the target to the image. Which result is useful depends on the proposed representation.