What I am trying to understand is shown in all three properties of a secure hash function, I will focus on Preimage attack resistance.
Preimage resistance: given a hash $h$, it's difficult to find $m$ s.t. $H(m)=h$.
My question is, what does "given" mean in this case?
Does it mean for a randomly selected $h$ from the output space of $H$, or can it be a specific hash selected by the adversary?
For example, assume a secure hash function $H$ defined as $H: \{0,1\}^*\rightarrow\{0,1\}^n$. We will build a hash function $H'$ that returns $0^n$ when input=0 otherwise returns $H(input)$.
An adversary in that example knows that if the $h=0^n$ then the message could be 0. So in the case that he can select the $h$ he can violate pre-image resistance, otherwise, in the case, the input is randomly selected, since $H'$ is secure, the probability of getting $0^n$ would be negligible and so the property still holds.
The same goes when selecting two messages $m_1,m_2$ for 2nd preimage, and collision resistance.
----------------------EDIT------------------------------
The link @hamidreza posted was really helpful. In the paper, they describe 3 definitions for preimage resistance, called Pre, e(everywhere)Pre, and an (always)Pre.
Pre: in which the adversary is given a hash of a randomly generated message along with the key (randomly generated as well) of the function.
ePre: I didn't quite get it, I think the adversary has $|Y|$ shots to generate a random message and its hash be included in the given set of hashes $Y$ for a specific key (randomly generated).
aPre: similar to the first one but stronger, the adversary can find a key for which he will be able to generate a collision for a hash of a randomly generated message.
So, to answer my question, the adversary doesn't get to choose the hash for any of the suggested definitions. Based on Pre at least, the $H'$ that I defined is pre-image resistant since the chance of picking $m=0$, or any other $m$ s.t. $H(m)==0^n$ is negligible.
Are the definitions correct?
