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Is my logic is correct?

$f:U(n)\rightarrow U(1)$ defined by $f(A)=detA$ is a group homomorphism so that induces $f^{*}: \pi_1(U(n))\rightarrow \pi_1(U(1))$ will be an Isomorphism right(I am not sure)? as $\pi_1(U(1))=\mathbb{Z}$ as $U(1)=S^1$ so $\pi_1(U(n))=\mathbb{Z}$

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Why is it an isomorphism? –  user38268 Oct 27 '12 at 12:15
    
Well, I just got an counter example that it is not true, $\mathbb{R}\rightarrow S^1$, $x\mapsto e^{ix}$ –  Bunuelian Trick Oct 27 '12 at 12:17
    
Well, it actually is an isomorphism in this case, but you're right that it isn't in general. –  Jason DeVito Oct 27 '12 at 12:39
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Have you tried using the long exact sequence of homotopy groups for $U(n-1)\hookrightarrow U(n)\to U(n)/U(n-1)$? –  Neal Oct 27 '12 at 13:41
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up vote 5 down vote accepted

As you saw by your counter example, a Lie group homomorphism does not induce an isomorphism of fundamental groups. But one way to use homomorphisms to determine fundamental groups is through the fact that if $G$ and $H$ are connected with $G$ simply connected and $G \to H$ is a surjective homomorphism with a discrete kernel $K$ contained in the center of $G$, then this map is a covering and the fundamental group of $H$ is isomorphic to $K$. So if you know that $SU(n)$ is simply connected then you can consider the homomorphism $$ SU(n) \times \mathbb R \to U(n), ~~ (A, t) \mapsto e^{it} A. $$ This is surjective with kernel isomorphic to $\mathbb Z$ so that $\pi_1(U(n)) \simeq \mathbb Z$.

Though I guess the easiest way to see that $SU(n)$ is simply connected is to use Neal's suggestion of applying the LES in homotopy associated to the fibration $$SU(n-1) \to SU(n) \to SU(n)/SU(n-1) \simeq S^{2n-1}.$$

But then you might as well compute $\pi_1 U(n)$ directly from the similar fibration $$U(n-1) \to U(n) \to U(n)/U(n-1) \simeq S^{2n-1}.$$

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