Preimage of a constructible set in spectrum of a subring

While working through a proof of this paper, at the beginning of page 42, the author seems to claim the following is true:

Let $$R\subset S$$ be rings, where $$R$$ is a finite type algebra over $$\mathbb F_p$$. Consider the associated map of the prime spectra $$\varphi:\text{Spec}(S)\rightarrow \text{Spec}(R).$$ Suppose that $$K\subset \text{Spec}(R)$$ is a constructible subset such that $$\varphi^{-1}(K)=\varnothing$$. Prove that there exists an $$R\subset R^{‘}\subset S$$, such that $$R’$$ is a finite type $$R-$$algebra and that if $$\psi:\text{Spec}(R’)\rightarrow \text{Spec}(R).$$ is the associated map of spectra, then $$\psi^{-1}(K)=\varnothing$$.

I believe that I have an argument for the case when $$K$$ is a finite subset. One could think of $$S$$ as a direct limit of its finitely generated $$R$$-subalgebras and therefore $$Spec(S)$$ should equal an inverse limit of the spectra of the finitely generated $$R$$-subalgebras. For each prime in $$K$$, choose a finitely generated $$R$$-subalgebra where it does not have a preimage, and the rest is clear. However, I don’t know what to do for the general case.

preimage of a torsion free subgroup

Let $$\phi: G \to H$$ be a surjective group homomorphism, and ker$$(\phi)$$ is torsion free. Let $$B$$ be a torsion-free subgroup of $$H$$. Show that $$A = \phi^{-1}(H)$$ is torsion free. I’m confused why we need the condition that ker$$\phi$$ is torsion free.

preimage resistance

I’m struggling to get a clear understanding of second preimage resistance and collision resistance.

Research on the internet yielded the following definitions:

Second pre-image resistance

Given an input m1, it should be difficult to find a different input m2 such that hash(m1) = hash(m2). Functions that lack this property are vulnerable to second-preimage attacks.

Collision resistance

It should be difficult to find two different messages m1 and m2 such that hash(m1) = hash(m2). Such a pair is called a cryptographic hash collision. This property is sometimes referred to as strong collision resistance. It requires a hash value at least twice as long as that required for pre-image resistance; otherwise collisions may be found by a birthday attack.

As far as I understand, every collision resistant hash function is also second pre-image resistant.

I don’t understand why collision resistance is harder to achieve, given that input m1 of second pre-image resistance could still be theoretically any input in the domain of the hash function.