Let $G$ be a connected reductive Lie group with Lie algebra $\mathfrak{g}$. That means that $\mathfrak{g}=Z_\mathfrak{g}\oplus[\mathfrak{g},\mathfrak{g}]$, and $[\mathfrak{g},\mathfrak{g}]$ is semisimple. Now, the centre $Z_G$ of $G$ is a closed normal subgroup, so $G/Z_G$ is a Lie group with Lie algebra $[\mathfrak{g},\mathfrak{g}]$. But the commutator subgroup $[G,G]$ is also a Lie group with Lie algebra $[\mathfrak{g},\mathfrak{g}]$.
Question: Is $[G,G]$ isomorphic to $G/Z_G$?
There is a natural map $$\pi:[G,G]\to G/Z_G$$ induced by the inclusion $[G,G]\subseteq G$, but I fail to see whether this is a Lie group isomorphism.
Edit 1: Actually, it's not an isomorphism according to Tobias Kildetoft's comment. But I would be happy with an affirmative answer to the following weaker statement.
Question': Is $\pi$ a covering of Lie groups?
Edit 2: We see that $\ker\pi=[G,G]\cap Z_G$ which is a closed normal subgroup with Lie algebra $[\mathfrak{g},\mathfrak{g}]\cap Z_{\mathfrak{g}}=\{0\}$, and hence is discrete. Thus, to show that $\pi$ is a covering of Lie groups, we would only have to show that it is surjective.
Question 2: Is $\pi$ surjective?