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\section{Pullback of Forms}
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\begin{definition}[Pullback]\label{def:pullback}
Let $f:N\to M$ be smooth and $\omega\in\Omega^{k}(M)$.
The \emph{pullback} $f^{*}\omega\in\Omega^{k}(N)$ is given by
\[
   (f^{*}\omega)_{q}
   := (df_{q})^{*}\,\omega_{f(q)}
   \quad\text{for }q\in N,
\]
where $(df_{q})^{*}:\mathrm{Alt}^{k}(T_{f(q)}M)\to\mathrm{Alt}^{k}(T_{q}N)$ is the
natural pullback along the linear map $df_{q}$.
\end{definition}

Given a differential form on the target manifold $M$, the pullback allows us to “transport” it back along the map $f$ to the domain $N$ in a consistent and smooth way. The pullback behaves contravariantly: the map $f$ goes forward, but the form gets pulled back.

The pullback generalizes the change-of-variable rule in integration, and it preserves the differential and algebraic structure of forms, including their degree and antisymmetry.

The last property in the wedge product theorem ensures
\(
   f^{*}(\alpha\wedge\beta)=f^{*}\alpha\wedge f^{*}\beta.
\)

This compatibility means that pullback preserves the wedge product structure, which is essential for computations involving forms under coordinate transformations and mappings between manifolds. This is particularly useful when computing integrals on curved spaces via charts or parametrizations.

\begin{example}[Pullback in Coordinates]
Let $f:\mathbb{R}^{2}\to\mathbb{R}^{2}$ be given by $f(u,v) = (x,y) = (u, uv)$ and consider the $1$-form $\omega = y\,dx$ on $\mathbb{R}^{2}$. Then
\[
  f^{*}\omega = f^{*}(y\,dx) = (uv)\,d(u) = uv\,du.
\]
Here, $f^{*}$ replaces $x$ and $y$ in terms of $u,v$, and applies the chain rule to the differential. The result is a new $1$-form defined on the domain $\mathbb{R}^{2}$ in $u,v$-coordinates.
\end{example}


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\section{Orientation}
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\subsection{Vector Spaces}

\begin{definition}[Orientation of a Vector Space]
Two ordered bases $\mathcal B=(v_{1},\dots,v_{n})$ and
$\mathcal C=(w_{1},\dots,w_{n})$ of $V$ are \emph{equivalent} if the
determinant of the change–of–basis matrix is positive.  
An \emph{orientation} on $V$ is an equivalence class of ordered bases.
\end{definition}

This definition captures the idea of “handedness.” For instance, in $\mathbb{R}^{3}$, a right-handed basis and a left-handed basis both span the space, but they are not smoothly deformable into one another without reversing orientation. Thus, every finite–dimensional real vector space has exactly two possible orientations: positive and negative.

\subsection{Manifolds}

On manifolds, the notion of orientation must be defined consistently across all tangent spaces in a smooth way.

\begin{definition}[Orientable Manifold]\label{def:orientable}
A smooth manifold $M$ is \emph{orientable} if there exists a choice of
orientation $\mathrm{or}_{p}$ on each tangent space $T_{p}M$ such that
for every chart $(U,h)$ the coordinate basis
$\bigl(\partial_{1},\dots,\partial_{n}\bigr)$ is positively oriented
relative to $\mathrm{or}_{p}$ for all $p\in U$.
\end{definition}

This ensures that orientation can be defined consistently over overlapping charts without contradiction. If this is not possible—if one must “flip orientation” to transition between some charts—then the manifold is not orientable.

\begin{theorem}[Equivalent Characterisations of Orientability]
For a smooth $n$–dimensional manifold $M$ the following are equivalent:
\begin{enumerate}
  \item $M$ is orientable in the sense of Definition~\ref{def:orientable}.
  \item There exists a nowhere–vanishing $n$–form $\omega\in\Omega^{n}(M)$
        (\emph{volume form}).
  \item $M$ admits an atlas $\{(U_{i},h_{i})\}$ such that for all
        overlaps the Jacobian determinants satisfy
        $\det\bigl(J_{h_{j}\circ h_{i}^{-1}}\bigr)>0$.
\end{enumerate}
\end{theorem}

These equivalent conditions provide different perspectives: (1) is geometric, (2) is analytic via differential forms, and (3) is coordinate-based. The existence of a global nowhere-vanishing top-degree form means one can integrate consistently with a fixed sign convention across the manifold.

\begin{example}[Orientability: Sphere vs. Möbius Band]
The sphere $S^{n}$ is orientable because one can consistently define an outward-pointing normal direction and a corresponding volume form.

For instance, on $S^{2}\subseteq\mathbb{R}^{3}$, the $2$–form
\[
  \omega = x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy
\]
restricted to $S^{2}$ is nowhere vanishing and gives the standard outward-oriented surface element. Thus, $S^2$ is orientable.

On the other hand, the Möbius band is not orientable: traveling once around the band reverses the local frame’s orientation. There is no globally consistent choice of orientation, and no nowhere-vanishing $2$-form exists on it.
\end{example}

Add some plots and give me the new latex code. The plots should make the concepts more intuitive. Plots should be entirely drawn in latex not python.