It is well known that the fundamental group of a pathconnected topological group is abelian. Suppose that $G$ is a connected topological group and let $Ab(G)$ the abelianization of the topological group $G$. Is there a relation between $\pi_{1}G$ and $\pi_{1}(Ab(G))$ ?

1$\begingroup$ There's a functorial homomorphism. Have you looked at any examples? Do you have any more precise question? $\endgroup$– YCorMar 25 '19 at 19:52

$\begingroup$ @YCor sure, how far the functorial homomorphism from being an isomorphism $\endgroup$– labMar 25 '19 at 19:59

1$\begingroup$ Have you looked at any examples? $\endgroup$– YCorMar 25 '19 at 20:06

$\begingroup$ @YCor no, do you have a trivial easy example ? $\endgroup$– labMar 25 '19 at 20:09

3$\begingroup$ SU(2) and SO(3) are good ones... $\endgroup$– David RobertsMar 25 '19 at 21:04
This question is arguably too broad, but I interpret that to mean the OP is somewhat new to this area. So, I'll give some basics, in a CW answer, and I encourage others to add to it if they want to. First, for any group $G$, the abelianization $Ab(G)$ is defined to be the group $G/[G,G]$ where $[G,G]$ is the commutator subgroup. There is a natural quotient map $G \to Ab(G)$. The assignment of $G$ to $Ab(G)$ is also a functor from the category of groups to the category of abelian groups, and this functor realizes the latter as a reflective subcategory of the former.
If $G$ is a topological group, then $G/[G,G]$ can be given the quotient topology. See Section 5 of these notes. A common situation of interest is when $G$ is a (compact) Lie group. Taking the fundamental group is a functor $X\mapsto \pi_1(X)$ to the category of groups. Note that $\pi_1(G)$ does not need to be a topological group. See the thesis of Jeremy Brazas. Because $\pi_1$ is a functor, there is a natural homomorphism $\pi_1(G) \to \pi_1(Ab(G))$. The comments demonstrate that this morphism need not be injective or surjective in general. Here are some links where examples are computed: here, here, here.
If $G$ is a compact connected Lie group, then standard classification results show that there is a diagram $G\xleftarrow{f}H\xrightarrow{g}T\times K$, where $f$ and $g$ are surjective with finite kernel (and so are coverings), and $T$ is a torus, and $K$ is a finite product of simple Lie groups, which have finite fundamental groups. This gives maps $\pi_1(G)\xleftarrow{\pi_1(f)}\pi_1(H)\xrightarrow{\pi_1(g)}\pi_1(T)\times\pi_1(K)$, where $\pi_1(f)$ and $\pi_1(g)$ are injective with finite cokernel. This shows that $\pi_1(G)$, $\pi_1(H)$ and $\pi_1(T)$ are the same up to a finite error. We also get surjective homomorphisms $\text{Ab}(G)\xleftarrow{\text{Ab}(f)}\text{Ab}(H)\xrightarrow{\text{Ab}(g)}\text{Ab}(T\times K)=T$, where $\text{Ab}(G)$ and $\text{Ab}(H)$ are compact connected abelian Lie groups and therefore tori. There are a few more details to sort out, but I am pretty sure that $\text{Ab}(f)$ and $\text{Ab}(g)$ are again finite coverings so $\pi_1(\text{Ab}(G))$ and $\pi_1(\text{Ab}(H))$ are again the same as $\pi_1(T)$ up to a finite error. For any given $G$ it should not be hard to pin down the details and find $\text{Ab}(G)$ and $\pi_1(\text{Ab}(G))$ explicitly.
As a consequence of the PeterWeyl Theorem, an arbitrary compact Hausdorff group can be written as the inverse limit of a filtered diagram of compact Lie groups, and one could use this to transfer some results from the Lie case to the general case.
For any connected Lie group $G$, there is a maximal compact subgroup $G_0\leq G$ such that the inclusion $G_0\to G$ is a homotopy equivalence. One should be able to learn something from this, but I have not thought through the details.

2$\begingroup$ There's no obvious reduction from connected Lie groups to the easy case of compact connected Lie groups, because taking the abelianization does not commute with "passing to a maximal compact subgroup". To be more concrete, if $G=\mathrm{SL}_2(\mathbf{R})$ then abelianization induces $\mathbf{Z}\to 0$ on $\pi_1$, but for the maximal compact subgroups, it induces the identity of $\mathbf{Z}$. $\endgroup$– YCorMar 27 '19 at 10:35