In Baker A.'s book section 3.6 about the complexification of a real lie algebra, it briefly introduces one theorem:
If $\mathfrak{g'}$ and $\mathfrak{g''}$ are two complexifications of $\mathfrak{g}$ then there is an isomorphism of $\mathbb{C}$-Lie algebras $\mathfrak{g'}\rightarrow\mathfrak{g''}$ which extends the identity function on $\mathfrak{g}\subset\mathfrak{g'}$
I can't understand why he talks about the identity function and what "extend" means. Mean the domain $\mathfrak{g'}$ contains another domain $\mathfrak{g}$ ?
If $X,Y$ are sets, $Z\subset X$, and $f:Z\to Y$ is a function, $g:X\to Y$ extends $f$ (or is an extension of $f$) if $f(z)=g(z)$ for all $z\in Z$. In other words, if the restriction $g|_Z$ of $g$ to $Z$ is just $f$ again.
For a set $X$, the identity function $\text{id}_X$ on $X$ is the function $\text{id}_X:X\to X$ such that $\text{id}_X(x)=x$ for all $x\in X$.
By definition of complexification, if $\mathfrak{f}$ is a complexification of $\mathfrak{g}$, then $\mathfrak{g}\subset \mathfrak{f}$. So if $\mathfrak{f},\mathfrak{h}$ are two complexifications of $\mathfrak{g}$, we can think about the identity function $\textbf{id}_\mathfrak{g}$ which takes an input $g$ from $\mathfrak{g}\subset \mathfrak{f}$ and returns $\textbf{id}_\mathfrak{g}(g)=g\in \mathfrak{g}\subset \mathfrak{h}$. So this function maps a subset of $\mathfrak{f}$ into $\mathfrak{h}$. More specifically, it maps the subset $\mathfrak{g}$ of $\mathfrak{f}$ to the subset $\mathfrak{g}$ of $\mathfrak{h}$.
An extension of the identity $\text{id}_\mathfrak{g}$ in this context would be a function $J:\mathfrak{f}\to \mathfrak{h}$ such that for all $g\in \mathfrak{g}$, $J(g)=\textbf{id}_\mathfrak{g}(g)$. The author is saying that if for any two complexifications $\mathfrak{f},\mathfrak{h}$ of $\mathfrak{g}$, there is such a function which is also an isomorphism.
Let me also make a quick comment on what's really happening here. We can get the technical details, but what's the importance of this result? The idea is that it doesn't really matter how we complexify $\mathfrak{g}$, we get something which is structurally identical. Isomorphisms (in whatever category, in this case complex Lie algebras) are exactly those functions which are bijections that exactly preserve all of the structure your spaces have (in this case, all of the complex Lie algebra operations commute with $J$). So $\mathfrak{f}$, $\mathfrak{h}$ can be thought of as interchangeable, and $J$ offers us a "translation" of what element of $\mathfrak{h}$ should be interchanged with a given element of $\mathfrak{f}$. But more importantly, $J$ extends the identity, which means it $J(g)=g$ for all $g\in\mathfrak{g}$. So $J$ gives us a way of seeing that $\mathfrak{f}$ and $\mathfrak{h}$ are structurally identical in a way that exactly preserves the way that $\mathfrak{g}$ is situated inside these structures. So the structures $\mathfrak{f}$,$\mathfrak{h}$ are structurally identical, and $\mathfrak{g}$ is identical inside them.