Suppose $g,h:S^1\rightarrow M$ are embeddings of $S^1$ into a smooth oriented three manifold. Let $F:M\times [0,1]\rightarrow M$ be a smooth ambient isotopy between $g$ and $h$, which means that $F_0=Id_M$, $F_t$ is a diffeomorphism of $M$ for each $t\in [0,1]$, and $F_1\circ g=h$.
From such an isotopy, we obtain a smooth map from the annulus into $M$, $A:S^1\times[0,1]\rightarrow M$ defined by $A(x,t)=F_t(g(x))$. I'm curious about how such a map can fail to be an immersion. If $G:S^1\times [0,1]\rightarrow M\times [0,1]$ is the map $G(x,t)=(g(x),t)$, then $G_*$ has full rank since $g$ is an embedding. Now $A=F\circ G$ and using the chain rule, you can compute $A_*$, which is locally a $3\times 2$ matrix. The first column being a nonzero vector, which is the push forward $(F_t)_*\circ g_*(\frac{\partial}{\partial x})$, the second column contains the time derivatives $(\frac{\partial F_i}{\partial t})_i$ where $F_i$ are the component function of $F$. So if $A_*$ is not injective at a point, then the columns of $A_*$ are linearly dependent. We can rule out the case that $(\frac{\partial F_i}{\partial t})_i$ is zero by reparametrizing.
My question is this, is it possible to rule out the case of $A_*$ having linearly dependent columns, or at the very least, homotope $A$ keeping the boundary knots $g,h$ fixed, so that the resulting map is an immersion?
References are welcome.