Smooth variation and variation field of a curve

For each $t$, $s\mapsto \alpha (s,t)$ is a smooth transverse curve by [F1], so its derivative at zero lies in $T_{\alpha (0,t)}M=T_{\gamma (t)}M$. In local coordinates it has components ${\mathrm{\partial }}_s{\alpha }^i(0,t)$, which are smooth in $t$; hence [F2] makes $V$ a vector field along $\gamma$. On a common finite subdivision the same coordinate argument applies stripwise, and agreement of the continuous transverse curves at each seam makes the variation field continuous there under the stated definition.

Differentiating either constant endpoint curve of a fixed-endpoint variation gives $V(a)=V(b)=0$. For a moving endpoint there is no such conclusion, which is why neither endpoint term may be discarded. If $M$ is empty, no given curve exists. In dimension zero every transverse curve is locally constant and $V=0$; dimension one is literal. The central variation $\alpha (s,t)=\gamma (t)$ has zero field and shows the degenerate case. The interval endpoints use the local-extension convention in [F1], and only finite supplied subdivisions occur. No choice axiom is used.