$\require{AMScd}\newcommand{\C}{\mathscr C}\newcommand{\D}{\mathscr D}\DeclareMathOperator{\obj}{Obj}\DeclareMathOperator{\id}{id}$
Let $L\leq \mathfrak{gl}(n,k)$ be a matrix lie algebra. The transpose has an interesting interplay with the commutator:$$ [A^\intercal, B^\intercal] = A^\intercal B^\intercal - B^\intercal A^\intercal = -(AB - BA)^\intercal = -{[A,B]}^\intercal. $$ Thereby, it is not formally a lie-Algebra homomorphism, but it still does not seem to change very much: For instance, The transpose maps lie subalgebras to themselves. What would happen if we would allow for more things to be a “Lie-algebra homomorphism” in a way that we can consider homs “up to slight correction”?
Such a correction can be described as a natural transformation from some functor to itself, because what they need to fulfil is precisely that the family of morphisms can be „dragged“ through any function just like the minus sign does. In fact, in the case of vector spaces, the NT needs to leave every 1D subspace invariant, which means we are restricted to scalar multiples of the identity. TODO REWORD
DEF Let $U\colon \C\to \D$ be a faithful functor. Define $\C_\sim$ to be the category with objects $\obj(C)$ and morphisms $$ \C(X,Y) := \{Uf\,\alpha_Y \mid f\colon X\to Y, \alpha: U\to U\}, $$ where composition is inherited from $\D$.
LEM $\C_\sim$ is well-defined as a category, and contains $\C$ as a subcategory by means of the embedding of morphisms $$ f\mapsto Uf;\id_X $$
Proof Identities are given by $\id_{UX} = U(\id_X\id_X) = U\id_X\,U\id_X$ which is of the required form because $X\mapsto U\id_X$ is a natural transformation from $U$ to itself. To see that our morphisms remain valid under composition, note that given $f: X\to Y, g: Y\to Z$ and two natural transformations $\alpha, \beta: U\to U$, we have $$ Uf\,\alpha_Y\,Ug\,\beta_Z = UfUg \alpha_Z\beta_Z = U(fg) (\alpha\beta)_Z, $$ which is a valid representation of a morphism in $\C_\sim$ because $\alpha\beta$ is still natural. To practise drawing a commutative diagram, this is the flip we've done.$$\tag*{$\blacksquare$}$$
\begin{CD} UX @>Uf>> UY @>Ug>> UZ \\ @. @V{\alpha_Y}VV @V{\alpha_Y}VV \\ @. UY @>Ug>> UZ \\ @. @. @V{\beta_Z}VV \\ @. @. UZ \\ \end{CD}
In this section, we will examine the introductory example of $k$-lie algebras, $\operatorname{char} k=0$, equipped with the forgetful functor $U$ to $\underline{\operatorname{k-Vec}}$. Remember that in this case, the only natural transformations $U\to U$ are scalar multiples. Greek letters like $\lambda, \mu$ will denote elements of our field.
It is straightforward to check that $0$ is indeed a zero object.
Let $K, L_1, L_2$ be lie algebras, where $K$ is associated with two morhisms $\lambda_i f_i\colon K\to L_i$. Can we find a product object $P$?