Here is the question I want to solve:
Let $f: R \rightarrow S$ be a map of commutative rings. Show that, if $\mathfrak{p}$ is a prime ideal of $S,$ then the inverse image $f^{-1}(\mathfrak{p})$ is a prime ideal of $R$.
My question is:
I need $f$ to be a homomorphism in my proof, does anything in the question givens implies that my function $f$ is a homomorphism?
There is no hope for this to be true if $f$ is not a homomorphism. Typically in algebra, a "map" between some structures is expected to be a homomorphism preserving the expected properties. I don't think anyone is really interested in set maps between rings.
Anyways, here's an argument. Recall that an ideal is prime iff the quotient by that ideal is a domain. Consider then the composition $R \longrightarrow S \longrightarrow S/\frak p$ of $f$ and the quotient map. The kernel of this map is exactly $f^{-1}[\frak p]$, so the first isomorphism theorem implies the existence of an injective ring homomorphism $\overline{f}: R/f^{-1}[\mathfrak{p}] \longrightarrow S/\mathfrak{p}$. $\mathfrak{p} \subseteq S$ is prime by assumption, so $S/\frak p$ is a domain. As $\overline{f}$ is injective, $R/f^{-1}[\frak p]$ is also a domain, so $f^{-1}[\frak p]$ is indeed prime.