compactness and fundamental group of $SL(n,\mathbb{C})$

Question

could any one help me to prove the heading?

$SL(n,\mathbb{C})$ is closed I can prove only, what are the other tools I need?

$SL(n,\mathbb{C})$ connected?simply connected?

Answer 1

$SL_{n+1}(C)$ has the unitary group $U_n$ as a deformation retract (Gram-Schmidt) so the fundamental groups are the same. The unitary group is simply connected for $n\ge 2$ and if $n=1$ the fundamental group is Z (the fundamental group of the circle).

Answer 2

The case $n=1$ is trivial, so assume $n\ge 2$. It is easy to see that $SL(n,\mathbb{C})$ is not compact, just look at diagonal matrices with entries $\lambda, \lambda^{-1}, 1, \ldots,1$, and let $\lambda \to \infty$. It is equally easy to see that it is connected, just deform the Jordan normal form to the identity matrix. For the fundamental group you need a little more sophisticated tools, see the other answer.