homomorphism of Lie superalgebras 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?
 A: It is possible to talk about odd homomorphisms as you have defined them, though an odd homomorphism will necessarily vanish on the subspaces $[\mathfrak{g}_{\overline{0}},\mathfrak{g}_{\overline{0}}]$ and $[\mathfrak{g}_{\overline{1}},\mathfrak{g}_{\overline{1}}]$ (assuming you are working over a field of characteristic not equal to $2$). In many interesting cases, this guarantees that the homomorphism will vanish on all of $\mathfrak{g}_{\overline{0}}$.
To see why the homomorphism will vanish on $[\mathfrak{g}_{\overline{0}},\mathfrak{g}_{\overline{0}}]$ and $[\mathfrak{g}_{\overline{1}},\mathfrak{g}_{\overline{1}}]$, let $\mathfrak{g}$ and $\mathfrak{g'}$ be Lie superalgebras, let $x$ and $y$ be homogeneous elements in $\mathfrak{g}$, and let $i: \mathfrak{g} \rightarrow \mathfrak{g}'$ be an odd homomorphism in the sense you have defined. Denote the homogeneous degrees of $x$ and $y$ by $d(x)$ and $d(y)$. Then $[x,y] = -(-1)^{d(x) \cdot d(y)} [y,x]$, so
$$
i([x,y]) = -(-1)^{d(x) \cdot d(y)} i([y,x]).
$$
Now using the fact that $i$ is a homomorphism,
$$
i([y,x]) = [i(y),i(x)] = -(-1)^{d(i(x)) \cdot d(i(y))}[i(x),i(y)] = -(-1)^{d(i(x)) \cdot d(i(y))}i([x,y]).
$$
Finally,
$$(-1)^{d(i(x)) \cdot d(i(y))} = (-1)^{(d(x)+1)(d(y)+1)} = (-1)^{d(x)\cdot d(y)+d(x)+d(y)+1}$$
so it follows that
$$
i([x,y]) = (-1)^{d(x)+d(y)+1}i([x,y]).
$$
If $x$ and $y$ are both of the same parity, this implies that $i([x,y]) = -i([x,y])$, and hence (by the assumption that $1 \neq -1$ in the given field) that $i([x,y]) =0$.
