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The function $y = |x|$ is not a differentiable function from $\mathbb{R} \rightarrow \mathbb{R}$. But considering the graph of $y = |x|$ as a subspace of $\mathbb{R}^2$, we can endow this space with a smooth structure consisting of the single smooth chart which projects the graph down to the x-axis.

What I am confused about is how to reconcile these two seemingly inconsistent viewpoints. Can someone explain why $y = |x|$ is not a smooth function yet it can be made into a smooth manifold?

Frederic Chopin
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    The point is that we should not care about how the manifold is embedded into $\mathbb R$. If you take just the graph, forget about all the other points of $\mathbb R^2$, you can just "unfold" the bend and see it is (topologically) just a straight line. – Wojowu Sep 07 '18 at 16:41

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Yes, of course, one could endow this graph with a smooth structure. But then it is not a smooth submanifold of $\mathbb{R}^2$. In order for the graph to be a smooth submanifold, it needs a smooth structure which makes the topological embedding $$(\mathrm{graph})\hookrightarrow\mathbb{R}^2$$ a smooth embedding (that is to say, in this context, a smooth immersion). The smooth structure you suggest does not satisfy this condition, and, in fact, no other structure does either.

Amitai Yuval
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  • This is subtly false as stated: you can certainly make that topological embedding a smooth map, by letting the derivatives decay to 0 at the cusp. But you will never be able to find an immersion doing so (which, I assume, is precisely what you meant: "...which makes the topological embedding a smooth embedding", in the sense of a topological embedding that is also an immersion). –  Sep 07 '18 at 17:39
  • @MikeMiller You are right, of course. I edited my answer. Thanks! – Amitai Yuval Sep 07 '18 at 18:22