Intuition of a non-differentiable function graph as a differentiable manifold? A $d$-dimensional topological manifold $M$ is a topological space for which there exists continuous maps (charts) to $R^d$, such that every point $p \in M$ is mapped by at least one of those charts.
A differentiable manifold $M$ is a topological manifold with the additional requirement that its Atlas contains only charts whose transition maps are all differentiable.
I find this second definition (of differentiable manifolds) very strange, because it only makes reference to the charts in the Atlas of manifold $M$, but not to the manifold itself.
If I'm correct, it means we can construct a differentiable manifold out of a non-differentiable continuous function:

`Take the graph $M$ of a non-differentiable function (e.g. $f(x)=|x|$). This clearly is a $1$-dimensional topological manifold. Now we can simply take the Atlas that consists of the single chart map that maps every point on the graph to the $x \in \mathbb R$ that produced it. This clearly is a chart map, and it clearly has a chart transition map to itself that is differentiable.

So this means that manifolds that have "kinks" in them, like the graphs of non-differentiable functions, can still be differentiable manifolds.
Could even a function like the Weierstrass function be a differentiable manifold?
What does "differentiability" of a manifold intuitively mean, since it is not intuitively the same thing as the differentiability of a real function?
 A: This is a good question. The point is that when you look at the graph of $|x|$ using "topological glasses" (so you only care about its structure as a topological space), there is no difference between this space and $\mathbb{R}$. By turning it into a smooth manifold using your single chart, you have effectively declared that when you wear your "smooth glasses", the graph of $|x|$ looks like your regular picture of $\mathbb{R}$ (and indeed, it will be diffeomorphic to it).
Allow me to give you an analogy. You are probably used to think about the set $\mathbb{R}$ (among other things) as a one-dimensional topological manifold. Let $S = \mathbb{R} \setminus \mathbb{Q}$ and choose an arbitrary bijection $\phi \colon \mathbb{R} \rightarrow S$. Using $\phi$, we can define a topology $\tau$ on the set $S$ by declaring $U \subseteq S$ to be open iff $\phi^{-1}(U)$ is open in the standard topology of $\mathbb{R}$. Endowed with this topology, $(S,\tau)$ becomes a one-dimensional manifold homeomorphic to $\mathbb{R}$ via $\phi$. But wait a second! The set $S$ "doesn't look" like a one-dimensional manifold! In fact, it looks totally disconnected! Well, it isn't a one-dimensional manifold if you consider it as a subspace of $\mathbb{R}$ with the induced topology (so it already comes with the usual notion of closeness defined by the topology). However, nothing prevents you from "removing the topological glasses" and, treating $S$ as naked set, relabeling the points of $S$ and redefining the the topology to turn this "ugly" set into a nice manifold.
Regarding your question of intuition, a relevant notion that is closer to the standard mental picture of a space with no kinks is that of an embedded submanifold in $\mathbb{R}^n$. An embedded submanifold $S$ in $\mathbb{R}^n$ can be described locally as a graph of a smooth function (in the usual sense) so it doesn't have your kind of kinks and the graph of $|x|$ won't be an embedded submanifold of $\mathbb{R}^2$. An abstract manifold doesn't come with an embedding and without an embedding, it is difficult to talk about kinks but it is true that any manifold is diffeomorphic to an embedded submanifold of $\mathbb{R}^n$ so up to the natural notion of equivalence, you can think of a manifold as a space without kinks (just like you can think of the ugly set $S$ as the nice space $\mathbb{R}$).
