Section 16: Problem 4 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

A map $f:X\rightarrow Y$ is said to be an open map if for every open set $U$ of $X$ , the set $f(U)$ is open in $Y$ . Show that $\pi_{1}:X\times Y\rightarrow X$ and $\pi_{2}:X\times Y\rightarrow Y$ are open maps.

Let $U\subset X\times Y$ be an open set, and $x\in\pi_{1}(U)$ . Then there exists $y$ such that $x\times y\in U$ . Since $U$ is open, there is a basis set $A\times B$ in $U$ that contains $x\times y$ . Since it is a basis set, $A$ is open in $X$ . Moreover, $x\in A=\pi_{1}(A\times B)\subset\pi_{1}(U)$ . Therefore, $\pi_{1}(U)$ is open. Similarly for $\pi_{2}(U)$ .

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Section 16: Problem 4 Solution

Section 16: Problem 4 Solution