Problem with the definition of Work of a Force Field

Given a force field $$F:A\subseteq\mathbb{R^3}\to\mathbb{R^3}$$ where A is an open connected set and Given a regular curve $$\phi:[a,b]\subseteq\mathbb{R}\to\mathbb{R^3}$$ such that $$\phi\left(\left[a,b\right]\right)\subseteq A$$ my book defines the Work $$W$$ of $$F$$ to move a particle from $$\phi(a)$$ to $$\phi(b)$$ as: $$W=\int_\phi\langle F,T\rangle dS$$ Where $$\langle F(x),T(x)\rangle$$ is the scalar product between $$F(x)$$ and the tangent versor $$T(x)$$ to the curve at the point $$x\in\phi([a,b])$$. My problem with this definition is that we defined the tangent versor $$T(t)$$ to the curve at a point $$t\in[a,b]$$ as $$T(t)=\frac{\phi^{'}(t)}{\Vert\phi^{'}(t)\Vert}$$ so $$T$$ is a function that goes from $$[a,b]\subseteq\mathbb{R}$$ to $$\mathbb{R^3}$$

while to consider the line integral $$\int_\phi f dS$$ i should have that $$f$$ is a function defined in $$\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}$$ to $$\mathbb{R}$$

but if the curve is not injective i can't consider the function $$\langle F,T\rangle:\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}$$ to $$\mathbb{R}$$.

So i don't get if $$\int_\phi\langle F,T\rangle dS$$ is an abuse of notation to mean $$\int_a^b\langle F(\phi(t)),T(t)\rangle\Vert\phi(t)\Vert dt$$

Or if i need to suppose that $$\phi$$ is injective (or something similar like that $$\phi$$ is simple)

EDIT: the problem is that if $$\phi$$ is not injective i could have that for two different points $$t_0$$ and $$t_1$$ in [a,b] i have that $$\phi(t_0)=\phi(t_1)$$ and if $$T(t_0)\neq T(t_1)$$ i can't consider a function that for every $$x\in \phi ([a,b])$$ gives me the tangent versor to the curve $$\phi$$ at the point $$x$$. (because for the point $$\phi(t_0)=\phi(t_1)$$ i have two possible tangent versors to consider)

while to consider the line integral $$\int_\phi f dS$$ I should have that $$f$$ is a function defined in $$\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}$$ to $$\mathbb{R^3}$$
$$\int_\phi f dS$$ is perfectly defined for $$f : \mathbb R^3 \to \mathbb R$$ by $$\int_a^b f(\phi(t)) \Vert \phi^\prime(t) \Vert dt.$$
• So? The problem is that $\langle F,T\rangle$ is not a definable function unless $\phi$ is injective Jun 10, 2021 at 14:53
• Why do you say that? For any $t \in [a,b]$, $F(\phi(t))$ is well defined as well as $T(t)$. If you have $\phi(t_1) = \phi(t_2)$ ($\phi$ not injective), then $F(\phi(t_1)) = F(\phi(t_2))$ while $T(t_1)$ may be different to $T(t_2)$. Jun 10, 2021 at 15:15
• Exactly so i can't consider a function $L:x\in\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}\to L(x)\in\mathbb{R^3}$ that gives me the tangent versor $L(x)$ for each $x \in \phi\left(\left[a,b\right]\right)$ and so i can't consider $\int_\phi\langle F,L\rangle dS$ because i can't define $L$ (remember that i need that $\langle F,L\rangle$ is a function from $\mathbb{R^3}$ to $\mathbb{R^3}$ but i can't define $L$ from $\mathbb{R^3}$ to $\mathbb{R^3}$ do you get the problem? Jun 10, 2021 at 15:22
• Sorry, I still don't understand what you mean. You now speak of $L$ which is not defined in your question. Jun 10, 2021 at 16:01
• Do you agree that to define $\int_\phi f dS$ i need that f goes from $\mathbb{R^3}$ to $\mathbb{R^3}$ right? So to define $\int_\phi\langle F,T\rangle dS$ i would need that $\langle F,T\rangle$ goes from $\mathbb{R^3}$ to $\mathbb{R^3}$ but it's not the case since $T$ goes from [a,b] to $\mathbb{R^3}$ and $F$ goes from $\mathbb{R^3}$ to $\mathbb{R^3}$ Jun 10, 2021 at 16:05