Submanifold such that $M-A$ is simply connected then $M$ is simply connected

algebraic-topologydifferential-geometrydifferential-topology

I am trying to prove the following statement

Let $M$ be a manifold and $A$ a submanifold of codimension greater or equal to $2$. If $M-A$ is Simply connected then $M$ is simply connected.

Now I was able to see that $M$ is path connected. For the part where $\pi_1(M)=\{0\}$ is where I don't know what to do . Does anyone have any suggestions? Thanks in advance.

Best Answer

A formalization hint. Any loop $\gamma$ in $M$ can be approximated by a differential loop $\alpha$ transversal to $A$. Now: (i) approximation implies homotopy, and (ii) since $\dim(\alpha)+\dim(A)<\dim M$, transversality just means $\alpha\cap A=\varnothing$.

Two edits for the asking.

(i) One can suppose $M\subset\mathbb R^p$. Then any two maps $\gamma,\alpha:I\to M$ are homotopic in $\mathbb R^p$ by linear interpolation. Then if $\alpha$ is close enough to $\gamma$, you get that interpolation inside a tubular neighborhood $W$ of $M$ in $\mathbb R^p$, hence can retract it into $M$.

(ii) Density of transversality can be formulated in many ways, here it is enough the one that so-called “parametrized”. I think a good reference is Abraham-Robbin’s Transversal Mappings and flows. The formal statement is just what needed: any mapping can be approximated by a differential mapping transversal to any given submanifold.