mjhjmtu

I'm dealing with Simon's algorithm a bit and "stumbled" upon something called for the algorithm. It is said that if the period is $s = 0^n$, then it is an injective function, that is, a 1 to 1 function. How can you show that this is so?

Then I would be interested. Moreover, if that is not the case, so $s neq 0^n$, then why is it a 2 to 1 function?

This is basically the definition of the type of function that you apply Simon's algorithm to. You are required to have a function $f(x)$ such that
$$
f(x)=f(y)
$$
if and only if $xoplus y=0$ or $s$.

Hence, if $s$ is all zeros, the outcomes are all unique: if $f(x)=f(y)$ then $x=yoplus 000ldots 0$, but bitwise addition modulo with 0 doesn't change the bit values, so $x=y$.

On the other hand, if $s$ is non-zero, there are exactly two values that give the same value of $f(x)$ since $xoplus00ldots 0=x$, just leaving the distinct $y=xoplus s$.

To add, following a comment. I suspect we need to go further back and understand the notation better. There is a function $f(x)$. It accepts, as an argument, a sequence of $n$ bit values, which we write as a variable $x$ (we write $xin0,1^n$ as a shorthand for conveying it's made up of $n$ bit values). The answer is a sequence of $n$ bit values, which we write as $y=f(x)$. We are promised that $a$, also a sequence of $n$ bit values exists such that
$$
f(x)=f(xoplus a).
$$
The calculation $xoplus a$ has a very specific meaning; we take each bit of $x$ (call the $i^th$ bit $x_i$) and each bit of $x$ and return the sequence where they have been added together modulo 2:
$$
x_ioplus a_i=x_itext XOR a_i=x_i+a_itext mod 2=left{beginarraycc 0 & x_i=a_i \ 1 & x_ineq a_iendarrayright.
$$
So, for example
$$
00110oplus 01010=01100.
$$
A key feature of this bitwise addition function is that
$$
xoplus aoplus a=x.
$$

An example of a suitable $f(x)$ is
$$
beginarrayc
x & 000 & 001 & 010 & 011 & 100 & 101 & 110 & 111 \
f(x) & 010 & 011 & 000 & 001 & 010 & 011 & 000 & 001
endarray
$$
Here, you can see that each value of $f(x)$ is repeated exactly twice. So, we find two values of $x$ that give the same output, say $001$ and $101$. These correspond to values $x$ and $xoplus a$, so we can find $a$ with
$$
a=xoplus(xoplus a)=001oplus101=100.
$$
Then you can check for every $x$ that $f(x)=f(xoplus 100)$.