If a short exact sequence is split-exact, can we conclude it is left-split?
Motivation: I am asking this question because I want to prove that if every $R$-module is projective, then every $R$-module is injective. I have already proved that if every $R$-module is projective, then the short exact sequence is split-exact. I need to move from there to prove that since we have a short exact sequence, then every $R$-module is injective.
Thanks in advance.
As linked in the comments by Osama Ghani, the relevant result here is the splitting lemma. This lemma tells us that a sequence of $R$-modules $$ 0 \to A \xrightarrow{f} B \xrightarrow{g} C\to0 $$ is isomorphic to the trivial short exact sequence $$ 0 \to A \xrightarrow{\operatorname{incl}} A\oplus C \xrightarrow{\operatorname{proj}} C\to0 $$ if and only if there is either $(1)$ a retraction of $f$ or $(2)$ a section of $g$.
Now, if we fix $C$ on the right, this sequence being split-exact, i.e. every projection onto $C$ having a section, is equivalent to $C$ being projective. Similarly, fixing $A$, every injection from $A$ having a retraction is equivalent to $A$ being injective.
Now consider $A$ arbitrary and an injection $A\hookrightarrow B$. Letting $C=B/A$ we obtain a short exact sequence $$ 0 \to A \to B \to C\to 0\,. $$ If we assume that every $R$-module is projective, in particular $C$ will be projective. Hence the above sequence splits. However, this then also implies that we can find a retraction of $A\to B$, that is, $A$ is injective.