How to define a map on Leibniz algebras?

Question

Free Leibniz algebras are defined as follows:

Let $X$ be a set and $F(X)$ be a non associative algebra and on that let $I$ be two sided ideal generated by $[a,[b,c]]-[[a,b],c]-[[a,c],b]$ for $a,b,c \in F(X)$. Then $L(X)=F(X)/I$ is called free Leibniz algebra.

In the case of free Lie algebras we have an obvious map from free Lie algebra to a Lie algebra. Is a a map $\phi \colon L(X) \to L$, where $L$ is an arbitrary Leibniz algebra definable?

Answer 1

The composite $X \to F(X) \to F(X)/I = L(X)$ is called the canonical map from $X$ to $L(X)$. The free Leibniz algebra satisfies the following universal property:

Let $X$ be a set, let $L(X)$ be the free Leibniz algebra on $X$, and let $i \colon X \to L(X)$ be the canonical map. For every Leibniz algebra $L$ and every map $f \colon X \to L$, there exists a unique homomorphism of Leibniz algebras $φ \colon L(X) \to L$ with $φ ∘ i = f$.

This is basically the same universal property as for free Lie algebras, but with every instance of “Lie algebra” replaced by “Leibniz algebra”.