Let $f:A\to B$ be a homomorphism of rings, and let $I:=\operatorname{ker} f$.
If $f$ is surjective, $\operatorname{dim}A\ge\operatorname{dim} B$.
I'd say that this holds because a chain of prime ideals, in $A/I$, yields a chain of prime ideals in $A$ via contraction. Since the contraction induces a bijection in this case, the inclusions of the chain in $A$ remain strict.
If $f$ is integral and injective, $\operatorname{dim}B\ge\operatorname{dim} A$.
This should be a consequence of the going-up theorem, for any chain of prime ideals in $A$ yields a chain of prime ideals in $B$, and again the inclusions stay strict.
If $f$ is integral, $\operatorname{dim}B\ge\operatorname{dim} A$.
This I don't know. We have that $A/I\to B$ is an integral injection, so $\operatorname {dim} B\ge \operatorname{dim} A/I$. However also $\operatorname {dim} A\ge \operatorname {dim} A/I$, so I don't know how I could deduce the thesis. Is it actually true? (I took the exercise from a paper where I already found some typos). If it is, could you give me only a hint to continue?
If $f$ is integral and injective we have that $A≃f(A)⊂B$ is integral and thus $\dim A=\dim B$.
For similar reasons, in the third case we have $\dim B\le\dim A$.