2
$\begingroup$

This is my problem: Let $X$ and $Y$ be topological spaces with topologies $T_X$ and $T_Y$, respectively. Recall, a map $X \xrightarrow{f} Y$ is called open if $\forall U \in T_X, f(U) \in T_Y$. Verify that the projection map $X \times Y \xrightarrow{\pi_X} X$ is an open map. Your result should naturally generalize to the other projection map as well.

This is how I approached the problem: By definition let $S$ be the product topology given by $S=\{\pi_{X}^{-1}(U) \mid U \in T_X\} \cup \{\pi_{Y}^{-1}(V) \mid V \in T_Y\}$. Want to show $\forall A \in S, \pi_X (A) \in T_X$.

Note:

$A= \begin{cases} \pi_{X}^{-1}(U)=\{(x,y) \mid x \in U, y \in Y\} \\ \pi_{Y}^{-1} (V)=\{(x,y) \mid x \in X, y \in V \} \end{cases}$

If $A=\pi_{X}^{-1}(U), \pi_X(A)=U \in T_X$. If $A=\pi_{Y}^{-1}(V), \pi_X(A)=X \in T_X$. Therefore, $A \in T_X$ and $\pi_X$ is an open map. The proof follows similarly for $\pi_Y(B)=V \in T_Y$.

Is there any other way to approach this problem/ is my logic and solution correct?

$\endgroup$
  • 3
    $\begingroup$ I think showing the projection is open on basis elements of $X$ is enough. $\endgroup$ – user99680 Feb 14 '14 at 4:58
  • $\begingroup$ Your $S$ is not correct. $\endgroup$ – Seub Feb 14 '14 at 8:11
2
$\begingroup$

By definition, a non-empty subset of $X\times Y$ is open if and only if it's of the form: $$\bigcup_{i\in I} (U_i\times V_i)$$ where $U_i$ are non-empty open subsets of $X$, $V_i$ are open subsets of $Y$. Since $\pi_X$ is the projection on the first component, we have: \begin{align*} \pi_X\Big[\bigcup_{i\in I} (U_i\times V_i)\Big] &=\bigcup_{i\in I} \pi_X[U_i\times V_i]\\ &=\bigcup_{i\in I}U_i . \end{align*}

$\endgroup$
  • 1
    $\begingroup$ How did you use the surjectivity of $\pi_X$? $\endgroup$ – user437309 Jan 31 '18 at 19:29
  • $\begingroup$ Surjectivity is, in fact, unecessary. $\endgroup$ – Fabio Lucchini Jan 31 '18 at 20:12

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.