Question about definition of a smooth manifolds (do all transition maps have to be smooth)?

I need to clear up my confusion on the definition of a smooth manifold. So we say that $M$ is a smooth manifold (of dimension $n$), if $M$ is Hausdorff and if every $x \in M$ is contained in a neighborhood $U$ that's homeomorphic to an n-ball (the pair $\phi, U$ is called a chart), and if two such charts $\phi_1, U_1$, and $\phi_2, U_2$ overlap, then

$$\phi_2 \circ \phi_1^{-1} : \phi_1(U_1 \cap U_2) \to \phi_2(U_1 \cap U_2)$$

is a smooth map.

But I remember my professor proving that a certain space was a smooth manifold by merely finding an atlas (an open covering of the space by charts) such that the above holds. But according to the definition I wrote, this would be insufficient. Can anyone clear up my confusion?

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Why would that be insufficient? – Potato Oct 13 '12 at 17:43
Well, consider two atlases $A1$ and $A2$. It could be that all transition maps within $A1$ and within $A2$ are smooth, but if two charts (one from A1 and one from A2) overlap, then that transition map may not be smooth, and the definition I wrote(which may be wrong), indicates that any transition map must be smooth. – Jorge Oct 13 '12 at 17:47
A manifold is a pairing of a topological space with an atlas. Topological spaces can support many incompatible manifold structures. There is not always a canonical choice. Your professor has taken a topological space and given it a smooth structure, so you get a manifold. Whether or not it is compatible with some other atlas doesn't really matter. – Potato Oct 13 '12 at 17:50
My preferred definition of a smooth manifold is that of using sheaves. We consider a ringed space $(M, \mathcal{O}_M)$, where $M$ is a Hausdorff space and $\mathcal{O}_M$ is a subsheaf of the sheaf of germs of continuous functions on $M$. Let $V$ be an open subset of $\mathbb{R}^n$. Let $\mathcal{O}_V$ be the sheaf of germs of smooth functions on $V$. The ringed space $(V, \mathcal{O}_V)$ is called a model space. If every point $p$ of $M$ has an open neighborhood $U$ such that $(U, \mathcal{O}_M|U)$ is isomorphic to a model space, $(M, \mathcal{O}_M)$ is called a smooth manifold. – Makoto Kato Oct 13 '12 at 19:23

I think your definition of a manifold is missing something. A topological manifold is a Hausdorff space $M$ such that every $x\in M$ has a neighbourhood $U$ that is homeomorphic to some open subset of $\mathbb R^n$. (Usually we also assume paracompactness.) The pair $(\phi,U)$ where $\phi$ is the homeomorphism I mentioned is called a chart.
Now the main point: to specify a smooth structure on $M$ we have to specify which charts we consider to be smooth. So, a smooth manifold is a topological manifold together with a set of smooth charts, which is a subset of all charts. And it is these smooth charts that we require to be smoothly compatible, i.e. for any two smooth charts $(\phi,U)$, $(\psi,V)$, we require that $$\psi\circ\phi^{-1}:\phi(U\cap V)\to\psi(U\cap V)$$ is smooth. A collection of such smooth charts that covers $M$ is called a smooth atlas.