# Confusion about "contradiction" in surjectivity of exponential map for $SL_2(\mathbb{C})$

In Hall Chap. $$3$$ Example $$3.41$$, it is shown that the exponential map is not surjective for $$SL_2(\mathbb{C})$$, by showing that there exists no matrix $$X\in sl_2(\mathbb{C})$$ such that $$e^X = \begin{pmatrix} -1 & 1\\ 0 & -1\end{pmatrix}$$.

Hall also later goes on to show in Chap. $$3$$ Corollary $$3.47$$ that for a connected matrix Lie group, every element $$A\in G$$ can be written in the form $$A = e^{X_1}e^{X_2}\cdots e^{X_m}$$ for $$X_1, X_2, \cdots, X_m\in g$$, where $$g$$ is the Lie algebra of $$G$$.

Now this is where my confusion arises. According to the Baker-Campbell-Hausdorff (BCH) formula, we can write $$e^Xe^Y=e^Z$$, where $$Z = X + Y + \frac{1}{2}[X,Y]+\cdots$$. Now, $$sl_2(\mathbb{C})$$ contains all traceless matrices, and $$[X,Y]$$ is traceless for any $$X,Y$$. So,$$X,Y\in sl_2(\mathbb{C})$$ should imply that $$Z\in sl_2(\mathbb{C})$$ too. So, if we iteratiely apply the BCH formula to $$A = e^{X_1}e^{X_2}\cdots e^{X_m}$$ (which we can, as $$SL_2(\mathbb{C})$$ is connected), we should end up with $$A = e^U$$ for $$U\in sl_2(\mathbb{C})$$, which contradicts what was proved at the start.

So where am I going wrong?

The BCH formula doesn't converge for all $$X$$ and $$Y$$, so you can't apply it in all cases.