1
$\begingroup$

I have the following definition of the projection on product topology:

Let $\{X_i\}_{I\in I}$ be a family of topological spaces and for $j\in I$ let $p_j:\prod\limits_{I\in I}X_i\rightarrow X_j$ be the projection onto the $j$th factor i.e. $p_j((x_i)_{I\in I})=x_j$. Then

  • For any $j\in I$ the function $p_j$ is continuous.

  • A function $f:Y\rightarrow\prod\limits_{i\in I}X_i$ is continuous if and only if the composition $p_j\circ f:Y\rightarrow X_j$ is continuous $\forall j\in I$.

However, I'm looking for a proof of this. Everything I've looked at gave part 1. as defined in product topology, but I have yet to see a proof of it (I imagine part 2. follows).

If anyone could provide a proof, that would be great.

$\endgroup$
2
$\begingroup$

Take any open set $U_j\subset X_j$ then $p_j^{-1}(U_j)=X_1\times X_2\times \ldots X_{j-1}\times U_j\times X_{j+1}\times\ldots X_n\times \ldots $ which is open in $\prod_{j\in \Bbb N} X_j$

$\endgroup$
3
  • $\begingroup$ Why is this open? Is it because each $X_i$ is also open? $\endgroup$ – mrose Oct 26 '17 at 2:41
  • $\begingroup$ Yes any open set in $\prod X_j$ is of the form $U_1\times Y_2\times \ldots U_n\times X_{n+1}\times \ldots$ $\endgroup$ – Learnmore Oct 26 '17 at 5:25
  • $\begingroup$ i.e finitely many are open in $X_i$ and the rest are the whole spaces $\endgroup$ – Learnmore Oct 26 '17 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.