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.


1 Answer 1


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 $\prod_{j\in \Bbb N} X_j$

  • $\begingroup$ Why is this open? Is it because each $X_i$ is also open? $\endgroup$
    – mrose
    Oct 26, 2017 at 2:41
  • 1
    $\begingroup$ 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, 2017 at 5:25
  • $\begingroup$ i.e finitely many are open in $X_i$ and the rest are the whole spaces $\endgroup$
    – Learnmore
    Oct 26, 2017 at 5:26

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