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Let f and g be functions such that (g[itex]\circ[/itex]f)(x)=x for all x [itex]\epsilon[/itex]D(f) and (f[itex]\circ[/itex]g)(y)=y for all y [itex]\epsilon[/itex]D(g). Prove that a g = f^-1

Pf/

How would you go about starting this besides saying

Let f and g be functions such that (g[itex]\circ[/itex]f)(x)=x for all x [itex]\epsilon[/itex]D(f) and (f[itex]\circ[/itex]g)(y)=y for all y [itex]\epsilon[/itex]D(g).

isn't obvious that the functions would have to be the inverse of each other, How else could you get the identity? So how do you prove it, can I just say clearly it is.

Pf/

How would you go about starting this besides saying

Let f and g be functions such that (g[itex]\circ[/itex]f)(x)=x for all x [itex]\epsilon[/itex]D(f) and (f[itex]\circ[/itex]g)(y)=y for all y [itex]\epsilon[/itex]D(g).

isn't obvious that the functions would have to be the inverse of each other, How else could you get the identity? So how do you prove it, can I just say clearly it is.

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