I'm confused by the below example (see the picture). Here is what I don't understand.
In the picture $X$ looks like a circle, so let's assume for simplicity it is a circle.
Along $X_1$, the outward-pointing normal vector is $n_{(1,x)}=(1,0)\in T_1(I)\times T_x(X)$.
Probably I misunderstand the notation, but I don't see why is this the outward-pointing normal vector at $x\in X_1$. This notation suggests that $1\in T_1(I), 0\in T_x(X)$ where $x\in X_1$. Why does $0\in T_x(X)$? Why does $(1,0)$ is the vector shown in the picture? In this case (when I interpret $X$ (and hence $X_1$) as $S^1$, $S^1$ lies in $\mathbb R^3$, so shouldn't a tangent vector have 3 components?
Any ordered basis for $T_{(1,x)}(X_1)$ is $0\times \beta$, where $\beta$ is an ordered basis for $T_x(X)$.
This causes the same question. Now $(1,x)$ is considered as a point on $X_1$, but $X_1$ in this case lies in $\mathbb R^3$, why does it have only 2 components? Also, why does a basis of $T_{(1,x)}(X_1)$ has the form as specified?
There are further questions but I think the first step is to understand what I asked above.
