Unitary group and unit circle Let $U(n)$ denote the group of complex unitary matrices, let $S^1$ be the unit circle in the complex plane. Then the map
$$f:U(n)\to S^1,\quad f(A)=det(A)$$
is a group homomorphism and a submersion. 
It is straight forward to show that $f$ is a group homomorphism. For the second part, we have the derivative of $f$ as follows:
$$df_A(X)=det(A)tr(A^{-1}X)$$
I wonder how can we show that $df_A:T_AU(n)\to T_{f(A)}S^1$ is surjective. Please help
 A: Note that $T_{f(A)} \mathbb S^1$ is one dimensional. To show that $df_A$ is onto, one only needs to find a tangent vector $X$ such that $df_A(X)\neq 0$. (Then the image would be at least one dimensional and thus is onto).
Also, it suffices to check that when $A=I$. (Let $B(t)$ be a curve in $U(n)$ such that $B(0)= I$ and $\dot B(0)=X$. So represent a vector in $T_I U(n)$. Thus $AB(t)$ represent a vector in $T_AU(n)$ and $AX\in T_AU(n)$. Also,  
$$f(AB(t)) = f(A) f(B(t)) \Rightarrow df_A(AX) = det(A) df_I(X),$$
so $df_A$ is onto if and only if $df_I$ is.)
We can identify the tangent space as 
$$T_AU(n) = \{ X: XA^* + AX^* = 0\}. $$
Let $X$ be a matrix with only nonzero entry at $(1, 1)$, which is $\sqrt{-1}$. This is a tangent vector at identity and $df_I(X) = \sqrt{-1} \neq 0$. Note that the tangent space of $T_I\mathbb S^1$ is isomorphic to $\sqrt{-1}\mathbb R$. Thus $df_IX \neq 0$ and we are done.
A: Let $g(t)=diag(e^{it}, 1,\ldots,1)$. Then $\det(g(t))=e^{it}$, whose derivative at $t=0$ is non zero. Hence the derivative of $\det$ at the identity is non zero, thus surjective (since $\dim S^1=1$).Since $\det$ is an homomorphism, the derivative is suerjective at all points of $U(n)$. 
