In general $d \phi (X)$ is not defined $X$ being a vector field Let $\phi:M\to N$ be a smooth map between manifolds and $X$ a vector field on $M$. Moreover, suppose that $d\phi$ is injective at every point. I want to show that in general $d\phi (X)$ is not defined as a vector field on $N$. (Vector fields are not assumed to be smooth).
Consider the map $e: \mathbb R \to S^1$ defined by $e(t)=(cos(t),sin(t))$. $X$ is the identity map on $\mathbb R$. My calculations showed that this is such a counterexample. Is this correct?
( $de_t(a)= (-a\sin(t),a\cos(t))$ implying $de$ being injective, and $e(0)=e(2\pi)$ with $de(0)$ not equal to $de(2\pi)$.)
 A: This choice indeed gives a counterexample, but the explanation would benefit from some polishing, especially since the chosen vector field $X \in \Gamma(T \Bbb R)$, $X_s := t \partial_s$, as well as the use of the canonical isomorphisms $T_s \Bbb R \cong \Bbb R$, that might lead to confusion. (The vector field $X$ is sometimes called the Euler vector field.)
The chosen map is $$e : \Bbb R \to S^1 \subset \Bbb R^2, \qquad e : t \mapsto (\cos t, \sin t),$$ and so $$de_s(a \partial_t\vert_s) = (-a \sin s, a \cos s),$$ where we implicitly use the canonical identification $T_{e(s)} \Bbb R^2 \cong \Bbb R^2$. Like you say, $e(0) = e(2 \pi)$, but
$$de_0 (X_0) = de_0 (0 \partial_t\vert_0) = (0, 0) \in T_{(1, 0)} S^1$$ but
$$de_{2 \pi} (X_{2 \pi}) = de_{2 \pi} (2 \pi \partial_t \vert_{2 \pi}) = (0, 2 \pi) \in T_{(1, 0)} S^1 .$$
Thus $e$ and $X$ do not determine a preferred vector in $T_{(1, 0)} S^1$ and hence do not determine a preferred vector field on $S^1$.

As discussed in the comments, another way to show that pushing forward the vectors in a vector field $X \in \Gamma(TM)$ by a map $F : M \to N$ does not determine a vector field on $N$ is to consider any vector field $X$ and any nonsurjective map $F$. If $q \in N$ is not in the image of $F(M)$, the set $\{dF_p(X_p) : p \in M\}$  contains no vector in $T_q N$ and hence $F$ and $X$ do not specify a vector field on $N$.
