Projection map

**HTale** · February 11th 2009, 02:51 PM

I am stuck by this question,

"In general show that for all $j \in I$ there is a map

$\pi_j : \prod_{i \in I} X_i \rightarrow X_j$

which is the 'projection onto the j-th factor' ".

I'm stuck because I didn't even know you could ask this question; I'm all at sea about how to even do this!

Note, that $\pi_j : A_1 \times A_2 \times \cdots \times A_j \times \cdots \times A_n \rightarrow A_j$ , given by $\pi_j(a_1, a_2, \ldots, a_j, \ldots, a_n) = a_j$

**Moo** · February 12th 2009, 09:28 AM

Hello,

Originally Posted by HTale

I am stuck by this question,

"In general show that for all $j \in I$ there is a map

$\pi_j : \prod_{i \in I} X_i \rightarrow X_j$

which is the 'projection onto the j-th factor' ".

I'm stuck because I didn't even know you could ask this question; I'm all at sea about how to even do this!

Note, that $\pi_j : A_1 \times A_2 \times \cdots \times A_j \times \cdots \times A_n \rightarrow A_j$ , given by $\pi_j(a_1, a_2, \ldots, a_j, \ldots, a_n) = a_j$

I'm not sure I grabbed the problem correctly...

Let's assume $I=\{i_1,i_2,\dots,i_j,\dots \}$ (a countable set, otherwise it makes no sense to write the product)
Let $x \in \prod_{i \in I} X_i$ .
Then x can be written as a tuple :
$x=(x_{i_1},x_{i_2},x_{i_3},\dots,x_{i_j},\dots)$ , where $x_{i_k} \in X_{i_k}$ for any $i_k \in I$

$n \in I$ means that there exists $j$ such that $n=i_j$

And $\pi_n$ will yield the $i_j$ -th coordinate of x, that is $x_{i_j}$

So it's defined... ?

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