# Tangent map of constant map

I want to prove the well known result that for a map $f:X \to Y$ between smooth connected manifolds $X$ and $Y$ it follows that the tangent map $T_xf: T_xX \to T_{f(x)}Y$ is zero iff $f$ is a constant map.

I know how to prove this for derivations but now I want to prove this result using the definition of a tangent vector as an equivalence class of curves $\gamma: (-\epsilon, \epsilon) \to X$ such that $\gamma(0) = x$ and $D(\varphi \circ \gamma_1)(0) = D(\varphi \circ \gamma_2)(0)$.

Unfortunately I have no clue how to proceed here. The tangent map is defined as $T_xf([c]) = [f \circ c]$ for $[c] \in T_xX$.

• Note that the forward result is false unless $X$ is connected. The proof of $\implies$ is going to have to use something more global, anyhow, so shouldn't depend on your particular definition of tangent vector. – Ted Shifrin Dec 7 '16 at 15:49
• I assumed the manifold to be connected. I corrected this. – JDoe Dec 7 '16 at 16:35
• OK, so you surely are going to have to use the chain rule a few times to do this problem. Where do you get stuck? – Ted Shifrin Dec 7 '16 at 16:41