TITLE: Help needed in showing $SU(n)$ is a submanifold of $U(n)$
QUESTION [5 upvotes]: I have proved that $U(n)$, which is a group of unitary matrices, is a smooth manifold of real Dimension $n^2$. Now I am trying to show that $SU(n)$, which is a group of unitary matrices with determinant 1, is a smooth submanifold of $U(n)$ of real dimension $n^2-1$. However I am stuck at a step:

The idea is to use the theorem: Suppose $f:N^n \to M^m$ is a smooth map between smooth manifolds $N,M$ of dimension $n,m$ respectively. Then preimage of any regular value $y$ of $f$ is a regular submanifold of $M$ and is of dimension $n-m$. Here, $y$ is a regular value means that the rank of $f$ at every point in the preimage of $y$ has rank exactly $m$.
To use this Theorem, I considered the determinant map
$$ det: U(n) \to S^1:=\{z \in \mathbb{C}:|z|=1\}. $$ Now I want to show that
$1 \in S^1$ is a regular value of the map $det.$ That is, I have to show that derivative of $det$ at an arbitrary $g\in det^{-1}(1)=SU(n)$ has rank $1$. For an arbitrary $Y \in Mat_{n \times n}\mathbb{C}$, I get $$D_gdet(Y)=\lim_{t \to 0}\frac{det(g+tY)-det(g)}{t}=\lim_{t \to 0}\frac{det(g+tY)-1}{t}$$

I don`t know how should I proceed from here. How can I simplify the RHS?

REPLY [5 votes]: I don't know if you're in the context of Lie Groups. In this case you just need to invoke the Closed Subgroup Theorem which states that every closed subgroup of a Lie Group is a Lie Group, which also means by definition that is a submanifold. 
$SU(n)$ is a closed subgroup of $U(n)$ hence a submanifold. To see that is closed just consider the function determinant.