# homomorphism of Lie superalgebras

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In the book Shun-Jen Cheng, Weiqiang Wang Dualities and Representations of Lie Superalgebrasm. One founds the following definition(Definition 1.3):

Let $\mathfrak{g}$ and $\mathfrak{g'}$ be Lie superalgebras. A homomorphism of Lie superalgebras is an even linear map $f: \mathfrak{g} \rightarrow \mathfrak{g'}$ satisfying $$f([a,b])=[f(a),f(b)],~ a, b \in \mathfrak{g}. ~~~~(*)$$

Here is my question:

Must a homomorphism of Lie superalgebras be even?

Assume $\mathfrak{g}$ is a Lie superalgebra, $A$ is a trivial $\mathfrak{g}$-supermodule. Then $A$ can be viewed as a Lie superalgebra with the zero superbracket. Let $\mathfrak{g'}=\mathfrak{g}\oplus A$ with $\mathfrak{g'}_{\bar 0}=\mathfrak{g}_{\bar 0}+A_{\bar 1}$, $\mathfrak{g'}_{\bar 1}=\mathfrak{g}_{\bar 1}+A_{\bar 0}$. Then $\mathfrak{g'}$ is a Lie superalgebra. We can define embeeding map $i: A \rightarrow \mathfrak{g'}=\mathfrak{g}\oplus A$, with the degree of $i$ is ${\bar 1}$. The map $i$ satisfies $$[i(a),i(b)]=i([a,b]).$$

Can we say this $i$ is a homomorphism of Lie superalgebras?

This post imported from StackExchange MathOverflow at 2014-11-12 16:42 (UTC), posted by SE-user Ruby

Most categories of superthings should admit a notion of "odd morphism." In the simplest case of supervector spaces this is because supervector spaces can be enriched over themselves. This is analogous to how chain complexes can be enriched over themselves, and consequently there is a natural notion of morphism of degree $i$, not necessarily zero, between chain complexes.
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