Trying to follow the reasoning in Lee's Smooth Manifolds book that $Lie(GL(n, \mathbb{C}))$ is isomorphic in the category of Lie Algebras to $\mathfrak{gl}(n,\mathbb{C})$.
He ends up with the following commutative diagram
$\require{AMScd}$ \begin{CD} Lie(GL(n, \mathbb{C})) @>{\cong}>> \mathfrak{gl}(n, \mathbb{C})\\ @V{\beta_*}VV @V{\alpha}VV\\ \beta_*(Lie(GL(2n, \mathbb{R}))) @>{\cong}>> \alpha(\mathfrak{gl}(n, \mathbb{C})) \end{CD}
Where the maps $\beta_*$ and $\alpha$ are the standard injective lie algebra homorphism from $M(n, \mathbb{C}) \rightarrow M(2n, \mathbb{R})$.
We have proven earlier that the bottom map is a Lie Algebra isomorphism and that the vertical maps are also Lie Algebra isomporphisms.
He then concludes that since the the top map is a vector space isomorphism, it is also a Lie Algebra isomorphism.
I must be missing something obvious, because I am not seeing how this follows. For one, we haven't even shown that it is a Lie Algebra homomorphism.
You’ve written the top map as a composite of three Lie algebra isomorphisms, so it’s a Lie algebra isomorphism. Notice that this is not exactly an isomorphism between $GL(n\mathbb C)$ and $\mathfrak{gl}(n,\mathbb C)$, since the former is not in the category of Lie algebras. The application of the functor $Lie$ is doing something!