Vector fields along maps: I need another sanity check Consider the definition of a vector field along a smooth map $f: M \to N$ where $M,N$ are smooth manifolds:
A vector field along $f$ is a continuous map $W \colon M \to TN$ such that $W(m) \in T_{f(m)}N$ for each $m\in M$. 
Say, $T_xN$ is isomorphic to $\mathbb R^n$ at each $x$. So an element in $TN$ is of the form $(x,v) \in \mathbb R^n \times \mathbb R^n$. 
I'm worried that I don't really understand this seemingly simple definition. Concretely, it seems to easy to define a vector field along a given map $f$: just define $W(m)  := (f(m) , 0)$. 

What am I missing?

 A: You are correct, that is a vector field along $f$. As you seem to be having some difficulties with this concept, I have included a few examples and comments below.

Another way to state the condition $W(m) \in T_{f(m)}N$ is that $\pi\circ W = f$ where $\pi : TN \to N$ is the projection map. With this in mind, one can define a vector field along a smooth map $f : M \to N$ as a continuous map $W : M \to TN$ such that the following diagram commutes:
$$\begin{array}{cc}
& & TN\\
& \nearrow & \downarrow\\
M & \xrightarrow{f} & N
\end{array}$$
where the diagonal arrow is $W$ and the vertical arrow is $\pi$.

Let $V$ be a continuous vector field on $N$; that is, $V : N \to TN$ is continuous and $\pi\circ V = \operatorname{id}_N$. Now define $W := V\circ f$. As 
$$\pi\circ W = \pi\circ(V\circ f) = (\pi\circ V)\circ f = \operatorname{id}_N\circ f = f,$$
we see that $W$ is a vector field along $f$. Your example corresponds to the case where $V$ is the zero vector field.

Not every vector field along $f$ is of the above form. Suppose $W$ is of this form, that is $W = V\circ f$. Then if $f(m_1) = f(m_2)$, we see that $W(m_1) = W(m_2)$. However, there can be vector fields along $f$ for which two points have the same image under $f$ but different images under $W$. For example, consider the map $f : \mathbb{R} \to \mathbb{R}^2$ given by $f(t) = (\sin t, \sin 2t)$. The image of this map is a figure eight:

Now consider the map $W : \mathbb{R} \to T\mathbb{R}^2 \cong \mathbb{R}^2\times\mathbb{R}^2$ given by 
$$W(t) = ((\sin t, \sin 2t), (\cos t, 2\cos 2t)).$$ 
Note, under the isomorphism $T\mathbb{R}^2 \cong\mathbb{R}^2\times\mathbb{R}^2$, the projection $\pi$ becomes projection onto the first factor which I will denote by $\operatorname{proj}_1$. As 
$$(\pi\circ W)(t) = \operatorname{proj}_1(W(t)) = \operatorname{proj}_1((\sin t, \sin 2t), (\cos t, 2\cos 2t)) = (\sin t, \sin 2t) = f(t),$$ 
$W$ is a vector field along $f$.
Note that $f(0) = f(\pi) = (0, 0)$ but $W(0) = ((0, 0), (1, 2))$ and $W(\pi) = ((0, 0), (-1, 2))$.

The example above associates to a point $m \in \mathbb{R}$ a vector $W(m)$ which is 'tangent' to $f(\mathbb{R})$ at $f(m)$. I write 'tangent' as $f(\mathbb{R})$ is not a manifold; what I really mean is $W(\mathbb{R}) \subseteq df(T\mathbb{R})$. It is worth noting that a vector field along $f : M \to N$ doesn't have to be 'tangent' to $f(M)$.
Let $f : S^1 \to \mathbb{R}^2$ be the inclusion map and define $W : S^1 \to T\mathbb{R}^2 \cong \mathbb{R}^2\times\mathbb{R}^2$ by $W(x) = (x, x)$. As before, under the isomorphism $T\mathbb{R}^2 \cong \mathbb{R}^2\times\mathbb{R}^2$, the map $\pi$ becomes $\operatorname{proj}_1$. As 
$$(\pi\circ W)(x) = \operatorname{proj}_1(W(x)) = \operatorname{proj}_1(x, x) = x = f(x),$$
$W$ is a vector field along $f$. In this case, $W$ is the outward unit normal vector field on $S^1$.
Note that in the verification that $W$ is a vector field along $f$, we did not refer at all to the second component of the map $W$. In fact, for any continuous map $\alpha : \mathbb{R}^2 \to \mathbb{R}^2$, the map $W : S^1 \to T\mathbb{R}^2\cong\mathbb{R}^2\times\mathbb{R}^2$ given by $W(x) = (x, \alpha(x))$ is a vector field along $f$. Furthermore, $W$ is actually of the form $V\circ f$ discussed above where where $V(x) = (x, \alpha(x))$.
A: Yes, defining $W$ to be the zero vector everywhere gives a vector field along $f$. But in general there are other interesting examples. Vector fields are far from unique, and in fact, they form an infinite dimensional vector space over $\Bbb R$. The one you picked is just the zero element of that vector space.
