Submanifold such that $M-A$ is simply connected then $M$ is simply connected 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.
 A: Hint: Given a closed curve $\gamma$ in $M$, problems can only occur if $\gamma$ intersects $A$, i.e., $Y:=\gamma^{-1}(A)$ is a non-empty subset of $[0,1]$. Now if we can deform $\gamma$ such that it avoids $A$, we are in $M-A$ and can contract. For the simplest case, consider an isolated point $t\in Y$. We want to "push" $\gamma(t)$ (i.e., continuously modify $\gamma$ in a small neighbourhood of $t$) in a direction that does not keep $\gamma(t)$ in $A$ and also not in a direction along $\gamma$ because that could "pull" other points of the curve into $A$. By the codimension condition, we have at least one "other" direction available to play with.

A: 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.
