[example: The Tangent Bundle of the Circle]
Let $S^1 \subset \mathbb{R}^2$ be the unit circle, and write $p=(\cos t,\sin t)$. The curve $c(t)=(\cos t,\sin t)$ has velocity
\begin{align*}
c'(t)=(-\sin t,\cos t).
\end{align*}
This vector is tangent to $S^1$ at $p$, and it is perpendicular to the radial vector $p$ because
\begin{align*}
(\cos t,\sin t)\cdot(-\sin t,\cos t)=-\cos t\sin t+\sin t\cos t=0.
\end{align*}
Thus the tangent line at $p$ is
\begin{align*}
T_pS^1=\{a(-\sin t,\cos t):a\in\mathbb{R}\}.
\end{align*}
The total space is
\begin{align*}
TS^1=\bigcup_{p\in S^1}\{p\}\times T_pS^1\subset S^1\times\mathbb{R}^2.
\end{align*}
Define
\begin{align*}
\Phi:S^1\times\mathbb{R}\to TS^1
\end{align*}
by
\begin{align*}
\Phi((\cos t,\sin t),a)=((\cos t,\sin t),a(-\sin t,\cos t)).
\end{align*}
If $t$ is replaced by $t+2\pi k$, then
\begin{align*}
\cos(t+2\pi k)=\cos t
\end{align*}
and
\begin{align*}
\sin(t+2\pi k)=\sin t,
\end{align*}
so the formula gives the same point of $TS^1$. Hence $\Phi$ is well-defined.
For each $p=(\cos t,\sin t)$, every vector in $T_pS^1$ has a unique form $a(-\sin t,\cos t)$, since $(-\sin t,\cos t)\neq(0,0)$. Therefore the inverse map is
\begin{align*}
\Phi^{-1}((p,a(-\sin t,\cos t)))=(p,a).
\end{align*}
So $TS^1$ is globally described as the product $S^1\times\mathbb{R}$. This example is special because the tangent bundle of the circle is globally a product; in general, a tangent bundle is only required to have such product descriptions locally.
[/example]