Intuition for Smooth Manifolds Consider the graphs of the functions $f_1(x) = |x|$, and $f_2(x) = x$ under the subspace topology of $\mathbb{R}^2$. 
Both of these graphs are smooth manifolds, just pick coordinate charts to be $(x, f_i(x)) \leftrightarrow x$. 
Moreover, they are diffeomorphic via the map $(x, f_1(x)) \rightarrow (x, f_2(x))$. 
This seems to clash with my intuition. For example, the graph of $f_1$ has a corner, so it "shouldn't" be smooth, much less diffeomorphic to $f_2$, which is just a straight line. 
Can someone explain what's going on here? 
In light of these examples, how should I visualize smooth manifolds and diffeomorphisms?
 A: You could do the same thing with the set $T=\{(x,g(x))\}$ for any continuous $g$.  The reason this seems non-intuitive is that you haven't used the smooth structure of $\mathbb{R}^2$ at all in defining the smooth structure of $T$; you've just taken the smooth structure on $\mathbb{R}$ and "transported" it onto $T$.  Another way to say it: The intuitive non-smoothness of $T$ (for, say, $g(x)=|x|$) comes from looking at the way that $T$ is sitting in $\mathbb{R}^2$.
Abstractly, it's very much the same as the following situation, which may be clearer. The integers $\mathbb{Z}$ form a group under addition.  The set $T = \{17,59\} \subset \mathbb{Z}$ is not a subgroup of $\mathbb{Z}$ under addition.  It is true that $T$ can be made into group by transporting the structure of a 2-element group onto $T$, but you don't expect that this group will have anything to do with $\mathbb{Z}$ as a group anymore since you didn't use the group structure on $\mathbb{Z}$ to define the group structure on $T$.  
A: Your intuition is broken because the inclusion of the graph of $f_1$ (with the smooth structure you describe) into $\mathbb{R}^2$ isn't smooth. The graph of $f_1$, as a subset of $\mathbb{R}^2$ with its usual smooth structure, is not a smooth manifold for exactly the intuitive reason. 
