I'm studying Differential Topology for a lecture, and we're using Guillemin & Pollack as the text for the lectures. In section 1.6, they define 'simply connected manifold' as a manifold $X$ in which every smooth map $f: S^1 \to X$ is homotopic to a constant map. However, the most general definition of fundamental group has as elements equivalence classes of continuous loops. Assuming both definitions are equivalent for manifolds, we would need some result like: every continuous loop in a manifold is homotopic to a smooth loop. Also, Sard's theorem implies that given any smooth map $f: X \to Y$, where $\dim{X} < \dim{Y}$, its image has measure 0. Considering strange continuous functions such as Peano curves, it seems even more confusing for me.

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    $\begingroup$ Yes, every continuous map between manifolds can be homotoped to a smooth map. Given a metric on the target manifold, you can actually make this approximation as close as desired. Yes, Sard's theorem has the implication as stated. What's your question there? It's a quick proof that $\pi_k(X) = 0$ when $ k < \dim X$ $\endgroup$ Mar 31, 2020 at 23:49
  • $\begingroup$ question was if it is true that every map is homotopic to a smooth map. I was wondering if you could indicate to me some sketch of a proof of this fact. $\endgroup$ Mar 31, 2020 at 23:53
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    $\begingroup$ Rob's answer has a sketch for maps out of 1-manifolds. In general, you can do similar stuff with charts on a manifold, and using bump functions in each chart! $\endgroup$ Mar 31, 2020 at 23:56
  • $\begingroup$ @OsamaGhani: I think the question is only about loops (i.e., maps out of 1-manifolds). $\endgroup$
    – Rob Arthan
    Apr 1, 2020 at 0:35

1 Answer 1


Any continuous path $p : [0, 1] \to \Bbb{R}^n$ is homotopic to the straight line path from $p(0)$ to $p(1)$. Hence any continuous path $q : [0, 1] \to X$ where $X$ is a smooth manifold can be subdivided into a finite sequence of paths $q_i : [t_i, t_{i+1}] \to X$, $i = 1, \ldots k$ that are all smooth. Now by looking at a chart around each $t_i$ with $1 < i < k$ you can adjust the $q_i$ to make their union smooth everywhere.


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