Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is my logic correct?

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

share|cite|improve this question
Why is it an isomorphism? – user38268 Oct 27 '12 at 12:15
Well, I just got an counter example that it is not true, $\mathbb{R}\rightarrow S^1$, $x\mapsto e^{ix}$ – Un Chien Andalou 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
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
up vote 6 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 $$U(n-1) \to U(n) \to U(n)/U(n-1) \simeq S^{2n-1}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.