1. ## Projections

Hi guys,

The question I'm doing states,

"Let $i_1, i_2, \ldots, i_n$ be indices in $I$, and assume that for each $i \in I$, we are given $U_i \subseteq X_i$. Show that $\bigcap_{1\leq r \leq n} \pi^{-1}_{i_r}(U_{i_r})$ can be described as a product of subsets of $X_i$'s"

I just want to see if my thinking here is correct.

We have that $\pi^{-1}_{i}$ is given by the projection onto the ith factor. So, $\pi^{-1}_{i_1} : U_{i_1} \times U_{i_2} \times \cdots \times U_{i_n} \rightarrow U_{i_1}$, for instance. We have the following fact for some $U_1\subseteq X_1, U_2 \subseteq X_2$,

$\pi^{-1}_1(U_1) = U_1 \times X_2$

where $\pi_1 : U_1 \times U_2 \rightarrow U_1$

Now, I know I have to use this to prove the above question, I'm just stuck on how I would represent $\pi^{-1}_{i_r}(U_{i_r})$ using the above, and how to then manipulate the intersections. There is a further property that

$\pi^{-1}_1(U_1) \cap \pi^{-1}_1(U_2) = U_1 \times U_2$,

which perhaps may come in useful, although I haven't seen any use for it.

HTale.

2. Originally Posted by HTale
$\pi^{-1}_{i_1} : U_{i_1} \times U_{i_2} \times \cdots \times U_{i_n} \rightarrow U_{i_1}$, for instance. We have the following fact for some $U_1\subseteq X_1, U_2 \subseteq X_2$

It should be
$\pi_{i_1} : U_{i_1} \times U_{i_2} \times \cdots \times U_{i_n} \rightarrow U_{i_1}$.

Let S be a subbasis for the product topology you mentioned. Then,

$S = \{\pi^{-1}_{i}(U_{i}) : U_{i}$ is open in $X_{i}, i \in I \}$.

Since the all finite intersection of elements of S forms a basis for (X, T),
$\bigcap_{1\leq r \leq n} \pi^{-1}_{i_r}(U_{i_r})$ is a basic open set for a topology T. If we denote a basic open set for a product topology T using a product form,

$\bigcap_{1\leq r \leq n} \pi^{-1}_{i_r}(U_{i_r}) = \prod _{i \in I} A_{i}$, where $A_{i} = U_{i}$ for $i = i_1, i_2, .., i_n$ in $I$ and $A_{i} = X_{i}$ otherwise.