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 $\forall A \in S, \pi_X (A) \in T_X$.


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

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?

  • 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

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*}

  • 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

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