Invertible functions

f : X → Y is invertible if there is a function g : Y → X with gof = I_X and fog = I_Y. Then g is the inverse of f, written f⁻¹.

f is invertible if and only if f is one-one and onto. So a function can be shown to be invertible by proving it bijective, without finding the inverse.

Both conditions are needed: if f were not onto, some y would have nothing to go back to; if f were not one-one, some y would have two candidates to go back to.

To find f⁻¹, write y = f(x) and solve for x. For f : R → R, f(x) = 5x − 2: y = 5x − 2 gives x = (y + 2)/5, so f⁻¹(y) = (y + 2)/5. Check: f⁻¹(f(x)) = (5x − 2 + 2)/5 = x and f(f⁻¹(y)) = (y + 2) − 2 = y.

Choosing the codomain as exactly the range makes a one-one map onto. f(x) = 4x + 3 from N to Y = {7, 11, 15, …} is invertible, with f⁻¹(y) = (y − 3)/4.

f⁻¹ undoes f; it is not the reciprocal 1/f. For f(x) = 5x − 2, f⁻¹(3) = 1 while 1/f(3) = 1/13.