I have the following problem:
Let $X$ be a compact oriented $n$-dimensional smooth manifold without boundary, and $Y$ be a compact orientable smooth submanifold without the boundary of codimension one. If the complement $X\backslash Y$ is connected, then the following hold:
There exists a smoothly embedded loop $\gamma\subset X$ intersecting $Y$ at one point, and the point of intersection is transverse.
We can find a trivial tubular neighborhood $Y\times [-1,1]\hookrightarrow X$. Let $p\in Y$ and consider the smooth embedding path $\alpha\colon [0,1/2]\hookrightarrow Y\times [-1,1]$ defined by $\alpha(t)=(p,2t-1)$. Since, $X\backslash Y\times [-1,1]$ is path connected we have a smooth embedding $\beta\colon [1/2,1]\hookrightarrow X\backslash Y\times [-1,1]$ from $(p,+1)$ to $(p,-1)$. Concatenating $\alpha$ and $\beta$ we have a continuous embedding $\widetilde \gamma\colon [0,1]\hookrightarrow X$ such that $\widetilde \gamma\big|(0,1)$ is an embedding and $\widetilde \gamma\cap Y=(p,0)$, a point of transverse intersection.
Question: To get $\gamma$ from $\widetilde \gamma$ we need to perturb $\widetilde \gamma$ so that $\widetilde \gamma$ remains the same everywhere except in the nbds of $(p,\pm 1)$; where we need to make $\widetilde \gamma$ smooth embedding. Is it possible?
So, considering the local charts ate $(p,\pm 1)$ the above problem reduces to the following problem:
Question: Let $f\colon(-1,0]\hookrightarrow \Bbb R^n$ and $g\colon [0,1)\hookrightarrow \Bbb R^n$ be two smooth embeddings such that $\text{im}(f)\cap\text{im}(g)=\big\{f(0)=g(0)\big\}$. Is there any smooth embedding $h\colon (-1,1)\hookrightarrow\Bbb R^n$ such that $h\big|(-1,-\varepsilon)=f\big|(-1,-\varepsilon)$ and $h\big|(\varepsilon,1)=g\big|(\varepsilon,1)$ for some $0<\varepsilon<1$.
Any help will be appreciated.