Question

Show that the inverse mapping f-1 of a mapping f exists iff f is one-one onto

Answer 1

Let f:A→Bf:A→B be a function. Suppose that the inverse function g:B→Ag:B→A is noninjective. Sso there exist b1≠b2∈Bb1≠b2∈B so that g(b1)=g(b2)g(b1)=g(b2). Then ff is not well defined since f∘g(b1)≠f∘g(b2)f∘g(b1)≠f∘g(b2), meaning that one element gets mapped to two different places.

This is a failure of the "vertical line test" in elementary calculus if that helps.