# Role of the test function $f$ in the definition of the tangent space at a point on a manifold

According to this video

On a manifold, $(\mathcal M)$, the tangent space is the set of all possible derivatives ("velocities") at a point $p\in \mathcal M$ associated with every possible curve $(\psi: \mathbb R \to \mathcal M)$ on the manifold running through $p.$ This can be seen as a set of maps from every curve crossing through $p,$ i.e. $C^\infty (t)\to \mathbb R,$ defined as the composition $\left(f \circ \psi \right )'(t)$, with $\psi$ denoting a curve (function from the real line to the surface of the manifold $\mathcal M$) running through the point $p,$ and depicted in red on the diagram below; and $f,$ representing a test function. The "iso-$f$" white contour lines map to the same point on the real line, and surround the point $p$.

The explanation involves the set of infinity curves $\psi(t)$ on a manifold going through a given point

$$\left\{\mathcal V_{\psi,p}\quad\vert \quad\psi\to X\right\}$$

and entails the derivative of their composition with a function $\color\red{f}:$ $(f\circ\psi)'(t)$ from the manifold to the real line:

My question is where in the Wikipedia definition of the tangent space (reproduced below) this function $f$ appears, its purpose, and if possible, whether an example can be provided to see its role.

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• Would Mathematics be a better home for this question? – Qmechanic Sep 9 '17 at 22:32

The function $f$ doesn't appear. Let's formalize some things. Two smooth curves $\gamma$ and $\lambda$ are tangent to zeroth order at $p\in M$ if $\gamma(t_0)=\lambda(t_0)=p$. Let these two curves be tangent to zeroth order. Let us define an equivalence relation on the set of all curves that are tangent to each other to zeroth order at $p$.
Two such curves are equivalent ($\gamma\sim\lambda$) if for any $f\in C^\infty(M)$ we have $$\frac{d(f\circ\gamma)}{dt}|_{t=t_0}=\frac{d(f\circ\lambda)}{dt}|_{t=t_0}.$$ A tangent vector at $p$ is then an equivalence class of this relation. Of course we need to do a lot more work than this, but I'm gonna omit all that. The point is that this definition of the equivalence relation is actually unnecessarily strong.
Let $(U,\varphi)$ be a chart so that $p\in U$. Then if $$\frac{d}{dt}(\varphi\circ\gamma)|_{t=t_0}=\frac{d}{dt}(\varphi\circ\lambda)|_{t=t_0},$$ then $\gamma$ and $\lambda$ are tangent to first order. There are two ways to look at this. Either just take away that it is enough for this condition to hold for all coordinate functions instead of an arbitrary function, or notice that $\varphi\circ\gamma$ is just a usual $\mathbb{R}\rightarrow\mathbb{R}^n$ curve, and its derivative is a well defined vector... in $\mathbb{R}^n$. This definition is independent of the chart used, so this essentially says two curves are tangent to first order if the euclidean tangent vectors of their coordinate representations agree in any chart.
This is why the function $f$ doesn't appear in the wikipedia article. The lecturer whose video you are watching did it via my first equivalence relation (derivatives of general functions agree), while the wikipedia description does it via the second equivalence relation (derivatives of coordinate functions agree).