$\newcommand{\qr}[1]{|#1\rangle}$In this lecture, it is nicely explained how to define an operator that computes a function $f(x)$. I know how to implement such operators. (We just define $O\qr{x}\qr{y} = \qr{x}\qr{y \oplus f(x)}$.)
However, it it said in the lecture that this effectively proves $O = O^\dagger$ and I fail to see it so clearly. It says $O = O^\dagger$ by construction. How can I see that so clearly as it is implied?
Showing that $O=O^\dagger$ is equivalent to showing that $O^2=\mathbb{I}$. In other words, $$ O^2|x\rangle|y\rangle=|x\rangle|y\rangle $$ for all $x$ and $y$.
To show this, we start from the definition of the oracle $$ O|x\rangle|y\rangle=|x\rangle|y\oplus f(x)\rangle $$ and apply $O$ again: $$ O^2|x\rangle|y\rangle=O|x\rangle|y\oplus f(x)\rangle=|x\rangle|y\oplus f(x)\oplus f(x)\rangle=|x\rangle|y\rangle $$ as required (since $a\oplus a=0$, and bitwise addition is associative).
Defining such oracles, you may visualize it as many controlled operations, especially $\text{CNOT}$s which is an easy way to build oracles.
We know the effect of the $\text{CNOT}$ is if the control is a 1 then we add 1 into the target (you can see it as part of a function itself but it is meant for one bit representing the output register). If we enumerate options on a simple 2-bit example with the first as control we have : $$\text{CNOT}(00) = 00; \text{CNOT}(01) = 01; \text{CNOT}(10) = 1(0+1)=11;\text{CNOT}(11) = 1(1+1)=10$$
We know also that we cancel the effect of the CNOT by applying it again. Take the action of a CNOT but now on images from a first CNOT: $$\text{CNOT}(00) = 00; \text{CNOT}(01) = 01; \text{CNOT}(11) = 1(1+1)=10;\text{CNOT}(10) = 1(0+1)=11$$
So you see that the effect on bits representing the output of your function represented by controlled operations.
The $ \oplus $ symbol illustrate that if I may say so.
