# continuous curve homotopic to smooth curve

Assume that $$\gamma : [0,1] \to M$$ is a continuous curve with initial and endpoint $$a$$ and $$b$$ respectively and $$M$$ is a differentiable manifold. The very well known Approximation Whitney's theorem states that is $$\gamma$$ is continuous and differentiable in $$A = \{0,1\}$$ we can find a smooth curve $$\sigma$$ joining $$a$$ and $$b$$ such that $$\gamma$$ and $$\sigma$$ be relative homotopic to $$A$$ (its extremes).

What happens if $$\gamma$$ is not smooth in $$A$$? I have though the next argument. Let $$(U,\varphi)$$ be a chart centered in $$\gamma(0) = a$$ such that $$\varphi(U)$$ is a disk. For simplicity I am going to assume that $$M$$ is 2-dimensional. If $$\gamma$$ was simple (without self-intersection) its easy to see that $$\gamma :[0,\varepsilon] \to \varphi(U)$$ is homotopic (relative to $$\gamma(0)$$ and $$\gamma(\varepsilon)$$) to the segment that joins both points.

What happens if its not possible to obtain a neighborhood of the point $$a$$ such that $$\gamma :[0,\varepsilon] \to \varphi(U)$$ is not simple? I mean the set of point self-intersection points of $$\gamma$$ could have $$a$$ like an accumulation point...

I'm confused. The following is true.

Suppose that that $$\gamma \colon [0,1] \to M$$ is continuous. Then $$\gamma$$ is homotopic to a smooth curve, relative to its endpoints.

So I don't understand your question. Here is a proof of the above.

First, use the (proof of the) Lebesgue covering lemma. The image $$\gamma([0, 1])$$ is compact, so we can cover it by a finite collection $$\{U_i\}$$ of charts. The preimage of $$U_i$$ is open so is a union of intervals. Taking components, we obtain an open cover of $$[0, 1]$$. Again appealing to compactness, we get a finite open cover of $$[0, 1]$$ subordinate to the cover $$\{ \gamma^{-1}(U_i) \}$$. Picking points we have a collection $$\{t_j\}_{j = 0}^n$$ so that

• $$t_0 = 0$$,
• $$t_j < t_{j+1}$$,
• $$\gamma([t_j, t_{j+1}])$$ lies in $$U_{i(j)}$$, and
• $$t_N = 1$$

Second, use straight line homotopy in charts to homotope $$\gamma$$, relative to the $$t_j$$, to a piecewise smooth curve.

Third and last, round the corners. Again, this is done in charts. QED.

Note that you can homotope $$\gamma$$ to be smooth this way, but not, say, analytic.