Why is $f:\mathbb{R}\to S^1$ $f(t)=(\cos(t),\sin(t))$ a local diffeomorphism? An example in my book says that $f:\mathbb{R}\to S^1$ defined by $f(t)=(\cos(t),\sin(t))$ is a local but not global diffeomorphism.
By the inverse function theorem, $f$ is a local diffeomorphism if the determinant of $df_x$ is nonzero.
I must be doing something wrong. Isn't
$$
df_x=\begin{bmatrix} -\sin x \\ \cos x\end{bmatrix}?
$$
I was expecting a square matrix in order to take the determinant.
 A: The map $f$ is smooth and the derivative at a point is an injective linear map. Restricting the codomain of derivative to the tangent space to a point on the circle, the derivative becomes a linear isomorphism.
Inverse function theorem of smooth manifold applies and you obtain that $f$ is really a local diffeomorphism.
Edit: To answer @Idonotknow's questions all at once.


*

*The map $f$ is not a global diffeomorphism for an obvious reason: the map is not injective as $f(x)=f(x+2\pi n)$ for all integers $n$ and all real numbers $x$.

*If you view the map as $f:\Bbb R\to\Bbb R^2$, the derivative at $x$ would be given by the linear transformation represented by the Jacobian matrix $$\begin{bmatrix}-\sin x\\\cos x\end{bmatrix}$$ just as what OP found.

*You check the above linear map's injectivity by letting it act on a nonzero vector $c\in\Bbb R$ (view it as a vector in the vector space $\Bbb R$), giving the nonzero vector $$c\begin{bmatrix}-\sin x\\\cos x\end{bmatrix}.$$ You check that is vector is nonzero, by calculating its inner product with itself: $$c\begin{bmatrix}-\sin x&\cos x\end{bmatrix}c\begin{bmatrix}-\sin x\\\cos x\end{bmatrix}=c^2\not=0.$$

*Viewing $f$ as a maps $f:\Bbb R\to\Bbb R^2$, the derivative at a point $x\in\Bbb R$ is a map between the tangent spaces $df_x:T_x\Bbb R\to T_{f(x)}\Bbb R^2$. What I mean by restricting the codomain, is that we choose the subspace $T_{f(x)}S^1$ of $T_{f(x)}\Bbb R^2$, and restrict the codomain to $T_{f(x)}S^1$. You can do so, because the image of $f$, by definition, is on $S^1$ only.
