I am trying to prove that the projection map $\pi_X:(X, T)\times (Y,J) \to X$ is an open map

But I don't know if I can use the basis element directly, so my proof is quite round about and lengthy


$\pi_X$ is an open map if for all open set $W \subseteq X \times Y, \pi_X(W)$ is open

We know that the basis of the product topology of $X,Y$ is $\mathcal{B} = \{U \times V|U \in T, V \in J\}$

Then the topology generated by the basis is $T_{XY} = \{\bigcup B| B \subset \mathcal{B}\}$

Then let $W$ be an open set, $W = \bigcup B$, for some collection $B$.

$\pi_X(\bigcup B) = \pi_X(\bigcup\{U\times V|U \in T, V \in J\}) = \bigcup\pi_x(\{U\times V|U \in T, V \in J\}) = \bigcup \{U\}$ which is open.

Can anyone help me identify where the proof can be made more rigorous or simpler? Ultimately, is it okay to just show this for the basic open sets rather than an open set?

  • 2
    $\begingroup$ Yes, it suffices to prove that the images of basic open sets are open, but at some point you need to prove this fact. It’s easy enough. Suppose that $\mathscr{B}$ is a base, and you know that $f[B]$ is open for each $B\in\mathscr{B}$. If $U$ is open, there is some $\mathscr{U}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$, and then $$f[U]=f\left[\bigcup\mathscr{U}\right] =\bigcup_{B\in\mathscr{U}}f[B]$$ is a union of open sets and hence open. $\endgroup$ – Brian M. Scott Jun 23 '16 at 1:49

The proof of your claim comes down to proving whether $\pi_X\left ( \bigcup_iW_i\right ) = \bigcup_i \pi_X(W_i)$ where the $W_i$ are product open sets. Clearly if $U\times V$ is a product open set in $X\times Y$ then $\pi_X(U\times V) = U$.

$(\Longrightarrow)$ Let $p \in \pi_X\left ( \bigcup_i U_i \times V_i\right )$, then there exists an element $q \in X \times Y$ such that $\pi_X(q) = p$. If we write $q = (q_X, q_Y)$, then $q_X = p$ and $q_X \in U_j$ for some $j$. Observe however, that we are able to pick $q_Y$ without loss of generality so that $q_Y \in V_j$ for the same fixed $j$, hence $q \in U_j \times V_j$, and therefore $p \in \pi_X(U_j \times V_j)$ so that clearly $p$ is contained in $\bigcup_i \pi_X(U_i \times V_i)$.

Containment in the other direction is simpler than the direction given above.


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.