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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?

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    I voted this question down because after your questions [here](http://math.stackexchange.com/q/214637/16627) and [here](http://math.stackexchange.com/q/222038/16627) no new ideas are required. Try harder before you ask!2012-10-27
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    @commenter thank you very much.2012-10-27

2 Answers 2

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$SL_{n+1}(C)$ 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).

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    I think you mean $n>2$ and $n=2$ instead of $n>1$ and $n=1$.2012-10-27
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    @LukasGeyer thanks2012-10-27
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    No. The unitary group $U(n)$ is never simply-connected (well, except for $U(0)$). It has fundamental group $\mathbb{Z}$ as shown for instance in http://math.stackexchange.com/questions/222038/fundamental-group-of-un2016-11-05
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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.