Let f: Z®Z be such that f(x) = x +1. Is f invertible? and if it is, what is its inverse?

1) The map $f$ is invertible if and only if $f$ is bijective if and only if $f$ is surjective and injective.

$f$ is surjective because $\forall y \in \mathbb{Z} \exists x \in \mathbb{Z}: f(x) = x + 1 = y (x = y - 1)$

$f$ is injective because $\forall x_1 \in \mathbb{Z}, \forall x_2 \in \mathbb{Z}, x_1 \neq x_2$$f(x_1) = x_1 + 1 \neq x_2 + 1 = f(x_2)$

Thus, $f$ is invertible.

2) The map $f ^{-1}: \mathbb{Z} \rightarrow \mathbb{Z}$ is inverse to $f$ if $f^{-1} f = f f^{-1} = e_{\mathbb{Z}}$ , where $e_{\mathbb{Z}}$ is the identical map $\mathbb{Z} \rightarrow \mathbb{Z}$ .

$f^{-1}(y) = y - 1$ because $f^{-1}f(x) = (x + 1) - 1 = x = \\ = (x - 1) + 1 = ff^{-1}(x)$