Mapping $\mathbb{Z}_k$ into $\{d,d+1,\ldots,d+k-1\}$ preserving value $\bmod k$ I want to give a (simple) map $f:\mathbb{Z}_k \to \{d,d+1,\ldots,d+k-1\}$ for $d,k \in \mathbb{N}$ and such that $\forall i \in \mathbb{Z}_k. f(i) \bmod  k = i$.
Is there a particularly simple way of doing this?
Goal
The goal is to formally prove that if $f$ is periodic with period $k$ then $\sum_{m}^{m+k-1} f \; l = \sum_{m+d}^{m+d+k-1} f \; l$. The above is the particular case $m = 0$.
Current solution (suggested by @kingW3)
$f(i) = (i-d) \; mod \; k + d$  and if $g(j) = j \; mod \; k$ then $g \circ f = Id$.
I'm interested in showing $f \circ g = Id$. How can I do it?

There's only one way to map $\Bbb{Z}_k$ onto a set of consecutive $k$ integer in such a way 
$$
f(\bar x)\equiv x\bmod k.
$$
That's because in any given set $K$ of $k$ consecutive integers there's only one $y\in D$ satisfying the displayed congruence for any given $\bar x$.
Thus, let $f(\bar x)=y$.
