Let $M$ be a smooth manifold, $\gamma:[0,a] \to M$ a smooth path. Assume we are given another smooth path $\phi:(-\epsilon,\epsilon) \to M$ that starts from an endpoint of $\gamma$, i.e $\phi(0)=\gamma(a)$.
Does there exists a smooth variation $f:(-\epsilon,\epsilon)\times [0,a]\rightarrow M$ of $\gamma$ such that $f(s,0)=\gamma(0)$ and $f(s,a)=\phi(s)$? (By variation I mean $f(0,t)=\gamma(t)$).
I am not insisting that the variation wille be defined on all $(-\epsilon,\epsilon)$, every variation which is defined for some "time" will be fine.
I think a problem might arise when a variation will have to pass through a "non-existent" region and so will not be smooth (Imagine a disk removed from the plane which lies inside an area where the variation is supposed to pass through. At some point there will be a "jump" - since each curve in the variation has a "discrete" choice: to pass above or below the forbidden region).
I think that allowing the variation to be defined on arbitrarily small intevals is supposed to overcome this, but I am unsure about how to do this.
Motivation:
In some scenarios, you want the endpoint of every curve in your variation to stay in a certain submanifold which intersects $\gamma(a)$. (See the comments here and here).
Remark:
The standard way of building smooth variations is different from what isrequired here;
For any given smooth variational field $V(t)∈T_{γ(t)}M$, we define $f(s,t)=exp_{γ(t)}(sV(t))$ where $exp$ is the exponential map w.r.t some Riemannian metric on $M$).
I don't know exactly, how general you want to situation to be. Under weak assumptions, which basically say that the image of $\phi$ stays away from $\gamma(0)$, I think this is easy to get as follows: The derivative $\phi'$ defines a vector field along the curve $\phi$. Choose open subsets $U\supset V\supset \phi((-\epsilon,\epsilon))$ such that $\gamma(0)\notin U$, $\bar U$ is compact $V\subset U$. (The assumption that $U$ can be chosen such that $\gamma(0)\notin U$ is exactly the restriction I was referring to in the beginning.)
Extend $\phi'$ arbitrarily to a local vector field defined on $U$ and then multiply by a bump function with support contained in $U$ which is identically $1$ on $V$. The result can be extended by zero to a globally defined vector field $\xi$ on $M$. By construction, this vector field has compact support and hence it defines a global flow, call that $\psi_t$. Now define the variation by $f:(-\epsilon,\epsilon)\times [0,a]\to M$ by $f(s,t):=\psi_s(\gamma(t))$. This is clearly smooth by the usual properties of flows and $f(0,t)=\gamma(t)$ for all $t\in [0,a]$. Also by construction $\xi(\gamma(0))=0$, so $f(s,0)=\gamma(0)$ for all $s$. Finally, along the image of $\phi$, $\xi$ coincides with $\phi'$ which means that $\phi$ is an integral curve of $\xi$. But this shows that $f(s,a)=\psi_s(\gamma(a))=\phi(s)$.