**Non-answer I:** just a reference for the difference between topological and abstract groups: I think the first part of Chapter I of the above Moore reference addresses the question, but I'm not sure it exactly answers it w/o more work. He notes at the start of chapter 1 that he is talking about abstract groups there, and "For topological groups [the abstract fundamental group] need not coincide with the usual [topological] fundamental group although it does in the most important cases (e.g. semi-simple Lie groups)", but I see no specific proof of this latter fact.

An abstract group $G$ is "simply connected" if for every central extension
$$1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$$
there is a *unique* (splitting) homomorphism $\phi: G\rightarrow E$ that composes with the quotient map to yield ${\rm id}_G$. He shows this is equivalent to $H^1(G,T)=H^2(G,T)=0$ where $T$ is the circle group, and notes in passing that $H^1(G,T)=0$ is equivalent to $[G,G]=G$.

A cover $E$ of $G$ (more properly defined on the morphism) is when $E=[E,E]$ and the kernel of the map from $E$ to $G$ is central in $E$. He then proves that any group with $G=[G,G]$ has a (unique) simply connected covering group $F$, and that this is characterized by the inflation property that $H^2(G,T)\rightarrow H^2(F,T)$ is the zero map. He then shows that that $F$ is universal in the sense expected. All of this is at the abstract group level from my reading.

Notably the "universal central extension" at the group level is the "universal covering group" at this level. In section 2 of Chapter I, he then compares to the topological case, and with Theorem 2.3, noting the ("somewhat amazing") fact that: "if
$$1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$$
is a topological extension of $G$ by $A$ which splits as extension of abstract groups, then it splits as extension of topological groups", assuming $G=[G,G]$ here (and not just that the derived subgroup is merely dense). In fact, he shows that a splitting map $\phi: G\rightarrow E$ must be continuous.

However, as he assumes that the extension is topological, it doesn't seem to show (from what I can tell) that the abstract universal covering group is indeed the topological universal covering group as would be desired.

**Answer II:** Another reference, more specific to $SL_n$ is Chapter 1 of http://users.ictp.it/~pub_off/lectures/lns023/Rehmann/Rehmann.pdf

Therein (first page) the central extension
$$1\rightarrow K_2(n,k)\rightarrow St_n(k)\rightarrow SL_n(k)\rightarrow 1$$
gives the Steinberg group as the universal central extension, and $K_2(n,k)$ as the (algebraic) fundamental group. There is a epimorphism surjecting $K_2(n,k)\rightarrow K_2(k)$, from the inductive limit $St(k)\rightarrow SL(k)$, and for $n>2$ this is isomorphic. Milnor notes that $K_2(R)$ is the direct sum of $Z/2$ and an uniquely divisible group, so in particular larger than the $Z$ from topological coverings. See Example 1.6 of "Algebraic K-theory and quadratic forms" http://www.springerlink.com/content/t025u1152j330325/ Note: I think he speaks of the Milnor $K$-group in general, but that it is not a bother at $K_2$.

In Theorem 1.2 of Rehmann's paper (and Theorem 4.2 for general Chevalley), the generators of the kernel are given, from Matsumoto's work.