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...


1 Answer 1


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.


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