Map for roots of a Lie group to roots of a special subalgebra? 
*

*For regular subalgebras $h$ of some group's Lie algebra $g$, $$ h \subset h $$ the root system of the subalgebra is a subset of the root system of the original's group algebra. 

*Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras
Nevertheless, there must be some map from the root system of the subalgebra to the root system of the original algebra because that's how embedding a subalgebra is defined. My problem is finding this map. 
In other words: Given a set of roots for the original algebra $g$ and a subset of this root system related in some way to the root system of the special subalgebra. This subset is, by definition of a special subalgebra, not directly the root system of the subalgebra, but there must be some map to the correct root system of the special subalgebra. How can I find the corresponding map?
 A: *

*As Tobias Kildetoft points out, the question does not make sense in this generality. Root systems are reasonably defined only for semisimple, or at best reductive, Lie algebras. But even a simple Lie algebra generally has tons of subalgebras which are far from being reductive. Or what would you say is "the root system of
$\mathfrak{h}:=\pmatrix{ * & \dots & * & * \\
0 & \ddots &  & * \\
0 & 0 & \ddots & * \\
0 & 0 & 0 & *} \subset \mathfrak{sl}_n(k)$"?


*Even if we restrict to semisimple subalgebras $\mathfrak{h} \subset \mathfrak{g}$, in general there is no reasonable map $R_\mathfrak{h} \rightarrow R_{\mathfrak g}$ between their respective root systems. For example look at the inclusion $\mathfrak{so}_5(\mathbb C) \subset \mathfrak{sl}_5(\mathbb C)$. Good luck finding a map from the root system $B_2$ to the root system $A_4$. So your statement

Nevertheless, there must be some map from the root system of the subalgebra to the root system of the original algebra because that's how embedding a subalgebra is defined.

must be based on a misunderstanding. Maybe you mean something the other way around, like "if there is an embedding of root systems $R_1 \hookrightarrow R_2$, and $\mathfrak{g}_1, \mathfrak{g}_2$ are split (say, complex) semisimple Lie algebras having the respective root systems, then there's an embedding $\mathfrak{g}_1 \hookrightarrow \mathfrak{g}_2$. (I'm not even sure if that is literally true, in any case it's non-trivial.)
But think of this: If you have an automorphism of $\mathfrak{g}$, then its fixed points form a subalgebra $\mathfrak{h}$ of $\mathfrak{g}$. In some nice examples this subalgebra is reductive or even semisimple. If the automorphism was chosen so as to stabilise a certain Cartan subalgebra, then one can get the root system of $\mathfrak{h}$ as (subsystem of) a "folding" (quotient) of the root system of $\mathfrak g$. Compare e.g. Semisimple complex Lie algebra of type $A_3$ contains a Lie subalgebra of type $B_2$ as fixed points of an automorphism., Folding and realization of type B Lie group as a subgroup of $GL_n$., Understanding $G_2$ inside Spin(7)? (EDIT: problem solved), and the comments to https://mathoverflow.net/a/244895/27465.
The main reason I bring up this class of examples is that here one does have a map between the root systems, but here it is the other way around: In the first question linked above, the inclusion $$\mathfrak{sp}_4(\mathbb C) \hookrightarrow \mathfrak{sl}_4(\mathbb C)$$
corresponds "contravariantly" to a quotient map $$A_3 \twoheadrightarrow C_2.$$
And sometimes it is even more intricate: "Folding" of $A_4$ gives the non-reduced root system $BC_2$, which has $B_2$ as a reduced subsystem. That's the connection of the root systems $B_2$ and $A_4$ that governs my above example $\mathfrak{so}_5(\mathbb C) \subset \mathfrak{sl}_5(\mathbb C)$:

$$\matrix{A_4 &\twoheadrightarrow & BC_2 \\
&& \cup \\
&& B_2}$$

You cannot express this with one map between $A_4$ and $B_2$ in either direction. So if there is any way in which one turns root systems into a category such that inclusions of semisimple Lie algebras would correspond to morphisms between root systems, these morphisms would not be maps in the standard sense: It's at least something like above, taking subsystems of quotients. (I am not even sure if that would "catch" all subalgebras. I do believe that in principle such an assignment would need to be contravariant though, i.e. to a morphism $\mathfrak{h} \rightarrow \mathfrak{g}$ would correspond a morphism $R_{\mathfrak{g}} \rightarrow R_{\mathfrak{h}}$.)


*On the positive side, one finds certain very special kinds (in particular, as you say "regular" -- a nomenclature which unfortunately is totally overused for c. 200 different things) of subalgebras by looking at (special kinds of) sub-root systems. That's the content of Borel-de Siebenthal and related theories ("Branching Rules"). You say you are interested in $E_6$, maybe this helps: https://mathoverflow.net/q/314025/27465. (On the other hand, folding gives something like a ghost of a group of type $F_4$ inside $E_6$ groups, cf. https://mathoverflow.net/q/203295/27465.)

