Does a break in a collision resistance property of a hash function by definition implies an attack at the first pre-image attack? Is there a formal security proof in the shape of reduction that states that if an attacker manages to break the collision resistance property of a cryptographic hash function (a random oracle) he will break the pre-image attack as well? To rephrase it: is first pre-image resistance as hard as collision resistance s.t if you break collision resistance you break first pre-image attack?
This relation obviously doesn't hold.

If you define "break" as faster than what's expected of an ideal hash function

Define TrivialCollisionHash = GoodHash(input.Skip(1 bit)). Finding pre-images for this is just as hard as for GoodHash, i.e. $2^n$. Finding a collision is trivial, just flip the first bit.

If you define "break" as faster than a certain computation bound

Use an ideal hash function. It has collision resistance of $2^{n/2}$ and first pre-image resistance of $2^n$. So an attacker with a computational bound between $2^{n/2}$ and $2^n$ can find collisions but not pre-images. For example an attacker who can perform $2^{160}$ oracle queries on a 256 bit hash function will most likely find a collision, but he probably won't find a pre-image.