Showing a set $Y \cong X_{\alpha}$ and that a projection map is an open map. Let $I$ be a set, $\{X_{\alpha}\}_{\alpha \in I}$ an indexed family of topological spaces and $X = \prod_{\alpha \in I} X_{\alpha}$. Fix $\alpha_0 \in I$ and for each $\alpha \in I - \{\alpha_0\}$, fix $x_{\alpha} \in X_{\alpha}$. For each $\alpha \in I - \{\alpha_0\}$, let $U_{\alpha} = \{x_\alpha\}$, and let $U_{\alpha_0} = X_{\alpha_0}$. Finally $Y \cong \prod_{\alpha \in I} U_{\alpha}$.
Show that in the subspace topology, $Y \cong X_{\alpha_0}$ and the projection map $X \to X_\alpha$ is an open map?

Can anyone give advice / assistance with this? This is a bit over my head with the amount of information in the problem statement.
 A: The idea is that you are identifying a subspace of $X$ which is homeomorphic to $X_{\alpha_0}$ by taking all sequences in $X$ which are constant everywhere except for the $\alpha_0$ coordinate.  As an example, consider $X=\mathbb{R}^3$ and the subspace $Y$ of all points $(x,0,0)$ with $x\in\mathbb{R}$.  Then it is fairly clear from experience in linear algebra that $Y\cong \mathbb{R}$, and this can be shown by using the projection map $p_1:Y\to \mathbb{R}$ onto the first coordinate; this is the homeomorphism.  The general case is very similar.
As for the projection $X\to X_\alpha$ being an open map, think about how the product topology on $X$ is even defined.  (Try taking a basis element of $X$ and projecting it into its $X_\alpha$ component. What is this set?)
A: To see $Y\cong X_{\alpha_0}$ consider the map $(x_\alpha)_{\alpha\in I}\mapsto x_{\alpha_0}$. Now, an open set it $Y$ is just a collection of tuples so that the components in the $\alpha_0$ slot form an open set in $X_{\alpha_0}$. So we have an inverse where we just map $x\in X_{\alpha_0}$ to the tuple with $x_\alpha$'s in the spots corresponding to $\alpha\in I-\{\alpha_0\}$ and $x$ in the $\alpha_0$ slot. These are two maps are clearly continuous, so we have a homeomorphism.
For the projections, what do you know about the product topology? The topology is the coarsest so that the projections are continuous. This basically means that the the topology is generated by sets of the form $\pi_\alpha^{-1}(U)$ for $U\subset X_\alpha$ open. With this we can see that the projection $\pi_\alpha$ are open as the image of any open set under a projection must be a union of finite intersections of open sets in $X_\alpha$.
