Question on showing if a topological space is a smooth manifold-show that coordinate charts are homeomorphic to $R^n$? [Sorry if the following is not rigorous enough as I am an undergrad physicist with no background in pure mathematics. Please correct me if I stated something wrong or imprecisely]  
In a course in Riemannian geometry, our professor gave us a definition of topological manifolds that (partly) had to do with the charts ${(U_a,\phi_a)}$ that covered the manifold and whose corresponding maps $\phi$ are homeomorphisms to $R^n$.  
The professor then proceeded on to defining a differentiable manifold using the usual condition that the transition maps $\phi_a^{-1} \circ \phi_
{\beta}(W)$ with $U_{\alpha} \cap U_{\beta}=W\ne \emptyset$ need to be diffeomorphisms(homeomorphisms and $C^{\infty}$).   
Now, in practice, to show that a topolgical manifold is smooth, we show the maps $\phi_{\alpha}$ are one-to-one, that the charts $(U_{\alpha},\phi_{\alpha})_{\alpha}$ cover the manifold and show that all $\phi_a^{-1} \circ \phi_
{\beta}(W)$ with $U_{\alpha} \cap U_{\beta}=W\ne \emptyset$ are diffeomorphisms.  
But, since a smooth manifold is also a topological manifold, then why don't we also have to show that all the $\phi_{\alpha}$ are homeomorphisms since this is a necessary condition to have a topological manifold?
If we show this, and also show that the transition maps are $C^{\infty}$ then the transition maps are diffeomorphisms, as we want to show. But, this does not always guarantee that the maps $\phi_{\alpha}$ are homeomorphisms.  
EDIT:
So, why should we just show that all the $\phi_{\alpha}$ are one-to-one? Why is it not needed to show that the coordinate maps $\phi_{\alpha}$ are onto and continuous with continuous inverse(since these along with the 1-1 condition tell us that they are homeomorphisms)? 
What I said above is also true when we want to show that a topological space is a smooth manifold. So, the question is even more general.  
[EDIT 2: I have edited the title due to a comment pointing out a mistake I have made. I switched from "topological manifold" to "topological space"]
 A: If you do not know that the $\phi_\alpha$ are continuous, then it does not suffice to just check that they are injective and the transition maps are diffeomorphisms.  For instance, let $M$ be any topological manifold and let $\phi:M\to\mathbb{R}^n$ be any bijection (not necessarily even continuous!).  Taking $\phi$ as our only chart on $M$, the only transition map is the identity $\mathbb{R}^n\to\mathbb{R}^n$ which is a diffeomrphism.  But this certainly doesn't make $M$ a smooth manifold.
On the other hand, if you know that the $\phi_\alpha$ are continuous (but not necessarily homeomorphisms) it suffices to check that they are injective by invariance of domain, a hard theorem in topology.  Specifically, invariance of domain says that if $U\subseteq\mathbb{R}^n$ is open and $f:U\to \mathbb{R}^n$ is a continuous injection, then $f$ is an open map.  
Now in your case, suppose we have $\varphi_\alpha:U_\alpha\to\mathbb{R}^n$ which is injective and continuous.  Since we are assuming we have a topological manifold (of dimension $n$), $U_\alpha$ is a union of open subsets that are homeomorphic to open subsets of $\mathbb{R}^n$.  By invariance of domain, $\varphi_\alpha$ is an open map when restricted to each of these open subsets, and it follows that $\varphi_\alpha$ is an open map on all of $U_\alpha$.  It is therefore a homeomorphism to its image.

It is crucial to this argument, though, that you already know you have a topological manifold.  If you just have some arbitrary space, there are easy counterexamples.  Indeed, similar to in the first paragraph, you can just take $M$ to be any topological space with a continuous bijection to $\mathbb{R}^n$ which is not a homeomorphism and use that as your only chart.  (Such a space is easy to construct; for instance, let $M=\mathbb{R}^n$ and your bijection be the identity, and give $M$ a topology by adding some new open sets to the standard topology on $\mathbb{R}^n$.)
