# Tranversal paths between two points

I have been studying differential topology from Hirsch, and sometimes in proofs he takes two points $$x,y\in M$$ a path between them and then just says that we can assume that this is transversal to a certain submanifold that we are interested in. Now I have tried to prove this myself but I don't see why this is true.

One attempt well let's take a path $$\gamma$$ between them such that $$\gamma(0)=x$$ and $$\gamma(1)=y$$, then using the fact that we have that transversal maps are dense we can approximate this map by transversal maps $$h_n:[0,1]\rightarrow M$$, but here I don't think I have any control over $$h_n(0)$$ and $$h_n(1)$$ and so this could be not $$x$$ and $$y$$.

Second attempt, we show that $$\{\gamma \in C^r([0,1],M) : \gamma(0)=x, \gamma(1)=y\}$$ is open in the strong topology, however this doesn't seem to be very true to me , since when we consider neighborhoods we lose control over the initial and final point again.

Another idea would be to consider the path an embedding which is possible, and then use the fact that $$C_S^r([0,1],\{0,1\},\gamma([0,1]),\{x,y\})$$ is dense, but here we would need the fact that the manifold that we want the map to be transverse too to be closed.

So I am out of ideas , and would appreciate some help with this. Thanks in advance.

Here is a useful corollary to the parametric transversality theorem (c.f. Guilleman and Pollack P.73):

Corollary. If, for $$f: X\to Y$$, the boundary map $$\partial f: \partial X\to Y$$ is transversal to $$Z$$, then there exists a map $$g: X \to Y$$ homotopic to $$f$$ such that $$\partial g = \partial f$$ and $$g\pitchfork Z$$.

Here $$\partial f = f|_{\partial X}$$. The full proof is a bit lengthy, so I won't write it out here, but the basic idea is to modify a homotopy $$F$$ of $$f$$ by a bump function which vanishes on neighborhoods of the boundary points. I don't have a copy of Hirsch unfortunately, but I imagine a similar result is discussed in his book.

Anyway, taking $$X=[0,1]$$ and $$f$$ to be some path with $$f(0)=x$$ and $$f(1)=y$$, then so long as $$\partial f:\{0,1\}\to Y$$ is transverse to $$Z$$ the result will follow. This is trivially true if $$x,y\notin Z$$ (which is the scenario in which this result would usually be applied).

I like your first attempt and I bet it's possible to make it or some variant of it work, but I think Hirsch wants you to get your hands dirty and work in coordinates since the assertion follows by standard arguments there. He probably doesn't give details because they're pretty tedious. The real content can be found in the following special case.

Locally, a codimension $$1$$ submanifold of an $$n$$-manifold looks like a hyperplane in $$\mathbf{R}^n$$. More precisely, for each $$p\in N$$, there is a submanifold chart $$(x,U)$$ about $$p$$ for which $$x(U\cap N)\subset \mathbf{R}^{n-1}\times\{0\}$$. So, in the special case that $$M=\mathbf{R}^n$$ and $$N\subset \mathbf{R}^{n-1}\times\{0\}$$ where $$x=(x^1,\ldots,x^n)$$ and $$y=(y^1,\ldots,y^n)$$ with $$x^n>0$$ and $$y^n<0$$, this can be accomplished by joining the smooth linear path $$\gamma_1\colon (x^1,\ldots,x^n)\to (x^1,\ldots,x^{n-1},y^{n})$$ (which is certainly transverse to $$N$$) to the smooth linear path $$\gamma_2\colon (x^1,\ldots,x^{n-1},y^{n})\to (y^1,\ldots,y^n)$$ at their common juncture. You'll need to smooth the corner and if you've never done this sort of argument before it's definitely an instructive exercise.

This special case is the real content and the rest is just fiddly details. Here's one way the general case can proceed.

Pick any continuous path $$\gamma$$ from $$x$$ to $$y$$ and give $$\mathop{\mathrm{Im}}\gamma$$ an open cover by charts such that every chart in the cover that intersects $$N$$ is a submanifold chart. You can do this by noting that every point not in the closure of $$N$$ has a coordinate neighborhood that does not intersect $$N$$. Compactness of $$\mathop{\mathrm{Im}}\gamma$$ allows you to assume this covering is finite and the argument now proceeds by induction, the base case being essentially what was done in the special case above.