(non?)-surjectivity of the exponential map to $SL(2,\mathbb{C})$

I need help figuring out what is wrong with the following proof that the exponential map is not surjective onto $SL(2,\mathbb{C})$. I have an exercise to prove that it IS surjective.

Given $M \in SL(2,\mathbb{C})$, assume $M = \exp(A)$ for some matrix $A$ with trace $0$. Let $J=PAP^{-1}$ be the Jordan normal form of $A$. Then $\exp(J) =\exp(PAP^{-1}) = PMP^{-1}$. $J$ has one of the two forms $\begin{pmatrix}\lambda & 0 \\ 0 & -\lambda\end{pmatrix}$ or $\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$. In the first case, $\exp(J) = \begin{pmatrix}e^\lambda & 0 \\ 0 & e^{-\lambda}\end{pmatrix}$ and in the second case, $\exp(J) = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$. Both of these are in Jordan form already, so we see that the Jordan form of $M$ is the exponential image of the Jordan form of $A$.

Now, this means that to find the full image of the exponential map, we must only find the matrices whose Jordan form is in one of the two above forms. But the matrix $\begin{pmatrix}-1 & 1 \\ 0 & -1\end{pmatrix}$ has neither of those two forms, and is in its unique Jordan form. Therefore, it cannot be in the image of the exponential map, either in $SL(2,\mathbb{C})$ or in $SL(2,\mathbb{R})$

Thanks

Your argument is correct. There must be something wrong with the exercise.

Exercise 4.4 of "Groups & Symmetries" by Kosmann-Schwarzbach asks you to prove the exponential is not surjective and your proof is both clever and correct.