Let $A$ be an $R$-Algebra. An $R$-linear derivation $d \colon A \to \Omega_{A/R}$ is called universal derivation or Kähler differential if for every $R$-linear derivation $D \colon A \to M$ there is a unique $A$-linear map $\Omega_{A/R} \to M$ making $$ A \xrightarrow{d} \Omega_{A/R} $$ $$ \searrow \hspace{.6cm} \downarrow$$ $$ \hspace{1cm} M $$ commute. I want to show that $d$ is always surjective and this is what I've tried:
It can easily be checked that $d \colon A \to \text{im}(d)$ is also a universal derivation. So $\Omega_{A/R}$ must be isomorphic to $\text{im}(d)$ with a unique isomorphism between them.
(We are not yet done since we want equality between $\text{im}(d)$ and $\Omega_{A/R}$. I tried plugging in $M = \text{im}(d)$ in the diagram above and use the uniqueness, but it didn't help)
By the construction of Kahler differential, we know that $\Omega_{A/R}$ can be seen as $F/E$, where $F$ is generated by symbols $da$,here $a\in A$, $E$ is generated by symbols $dr$($r\in R$) and $d(a_1+a_2)-da_1-da_2$ as well as $d(a_1a_2)-da_1 \cdot a_2-a_1\cdot da_2$ ($a_1,a_2\in A$). It's obvious that the map is surjective. Then $$\begin{array}{ccc}&&F/E \\ &\nearrow&\downarrow\\ A &\rightarrow & \Omega'_{A/R} \\ & \searrow &\downarrow \\ && F/E\end{array}$$ Thus $F/E \to \Omega'_{A/R}\to F/E$ is isomorphism. Which means $A\to \Omega'_{A/R}$ is surjective.