Let $R$ be a commutative ring and consider the pull-back diagram $$ M\times_L N \rightarrow M $$ $$ \hspace{5mm} \downarrow \hspace{15mm} \downarrow f $$ $$ \hspace{3mm} N \hspace{5mm} \xrightarrow{g} \hspace{3mm} L $$ of finitely generated $R$-modules. Taking symmetric algebras yields $$ \operatorname{Sym}\left(M\times_L N\right) \rightarrow \operatorname{Sym}(M) $$ $$ \hspace{10mm} \downarrow \hspace{30mm} \downarrow \operatorname{Sym}(f) $$ $$ \hspace{3mm} \operatorname{Sym}(N) \hspace{5mm} \xrightarrow{\operatorname{Sym}(g)} \hspace{3mm} \operatorname{Sym}(L) $$ which we can not in general expect to be a pull-back diagram.
My question is: When does it become a pull-back diagram?
(e.g. $R$ is a field, or $M$, $N$, $L$ are all free, or $f$, $g$ are injective, or $(f \ \ g)$ is surjective, etc.)
Is there a nice description between the symmetric algebra of a pull-back and the pull-back of symmetric algebras?
I tried with some examples where it became a pull-back diagram as follows: $$ \mathbb{k}\left<\beta\right> \rightarrow \mathbb{k}\left<\alpha,\beta\right> $$ $$ \hspace{1mm} \downarrow \hspace{20mm} \downarrow $$ $$ \hspace{3mm} \mathbb{k}\left<\beta,\gamma\right> \hspace{2mm} \rightarrow \hspace{3mm} \mathbb{k}\left<\alpha,\beta,\gamma\right> $$ is a pull-back diagram of $\mathbb{k}$-vector spaces and after taking symmetric algebras it becomes another pull-back diagram $$ \mathbb{k}\left[\beta\right] \rightarrow \mathbb{k}\left[\alpha,\beta\right] $$ $$ \hspace{1mm} \downarrow \hspace{20mm} \downarrow $$ $$ \hspace{3mm} \mathbb{k}\left[\beta,\gamma\right] \hspace{2mm} \rightarrow \hspace{3mm} \mathbb{k}\left[\alpha,\beta,\gamma\right] $$ but I failed to find a nice assumption in general. It would be helpful for me to hear any comments. Thank you.
