Let $\mathfrak H$ be the Heisenberg group and $\mathfrak h$ its lie algebra. Show that $\exp: \mathfrak h \to \mathfrak H$ is a bijection.
I was trying in that way that $\mathfrak H=\Bigg\{ \begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \\ \end{pmatrix}| x,y,z \in \Bbb R\Bigg\}$
then $Z(\mathfrak H)=\Bigg\{ \begin{pmatrix} 1 & 0 & z \\ 0 & 1 & o \\ 0 & 0 & 1 \\ \end{pmatrix}| z \in \Bbb R\Bigg\}$ and $\mathfrak H/(Z(\mathfrak H)) \cong S^1 \times S^1$ which is compact and connented. Moreover $Z(\mathfrak H) \cong \Bbb R$ is connected. I don't know the topological relations as much.
I can prove one fact that "If $G$ be a commutative connected matrix Lie group with Lie algebra $\mathfrak g$. Then the exponential map would be surjective. It comes from the fact that 'for $G$ be a connected matrix Lie group with Lie algebra $\mathfrak g$; then for any $X \in \mathfrak G$ $\exists X_1,...,X_r\in \mathfrak g$ s.t $X=e^{X_1}..e^{X_r}$.
Now if anyone would refer any theorem or something to prove it please give me some reference or prove it in the answer. Thanks a lot in advance.
In this case, the exponential map is just the map$$\begin{bmatrix}0&x&z\\0&0&y\\0&0&0\end{bmatrix}\mapsto\begin{bmatrix}1&x&z+\frac12xy\\0&1&y\\0&0&1\end{bmatrix}$$and it is quite easy to prove that this map is a bijection from $\mathfrak h$ onto $\mathfrak H$.