Two different morphism of $K$-schemes have different image points?

Question

If $K$ is a field and $X\rightarrow \operatorname{Spec}K$ a $K$-scheme, and I have two morphisms of $K$-schemes $f_1,f_2:\operatorname{Spec} K\rightarrow X$ that are different, does it follow that $f_1$ and $f_2$ have different image points? (I came along this question while trying to understand the proof of Görtz/Wedhorn Prop. 4.35.)

Answer 1

Question: "..does it follow that f1 and f2 have different image points?"

Answer: Let $X:=Spec(A)$ and let $\mathfrak{m}_i:=ker(f_i) \subseteq A$ be the kernel of $f_i$. By definition: the topological space of $Spec(A)$ is the set of prime ideals in $A$. If $f_1\neq f_1$ there is an element $x\in A$ with $f_1(x)\neq f_2(x)$. Let $a:=f_1(x)$. It follows $f_1(x-a)=0$ and $f_2(x-a)\neq 0$ hence $a\in \mathfrak{m}_1, a\notin \mathfrak{m}_2$ hence $\mathfrak{m}_1 \neq \mathfrak{m}_2$. By definition: If you view the maps $f_i$ as maps of schemes you get the following:

$$Im(f_i):=\{\mathfrak{m}_i\} \subseteq X$$

and it follows $Im(f_1) \neq Im(f_2)$.