The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.

Prove that if $X=X_1\times X_2$ is a product space, then the first coordinate projection is a quotient map.

Definition: Let $X$ and $Y$ be two topological spaces and $f:X→Y$ a surjective map. The map $f$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $f^{−1}(U)$ is open in $X$. Equivalently, a subset $U$ of $Y$ is closed in $Y$ if and only if $f^{−1}(U)$ is closed in $X$.

I was able to prove the projection mapping $\pi_1:X\rightarrow X_1$ was onto and continuous. I am having a bit of a pickle showing is $U \subset X_1$, $\pi_1^{−1}(U)$ open in $X \Rightarrow U$ open in $X_1$. I do have an idea to it.

Let $U_1\subset X_1$. Since $\pi_1$ is surjective, then $\pi_1(\pi_1^{-1}(U_1))=U_1, \forall U_1 \in X_1$. Consider the collection $\mathscr{B}_{X_1\times X_2}$ of all subsets of the form $O_1\times O_2$ where $O_1$ and $O_2$ are open in $X_1$ and $X_2$ respectfully. Thus $\mathscr{B}_{X_1\times X_2}$ is a basis. Let $W=\pi_1^{-1}(U_1)$ be an open subset of $X$ containing the point $x$. Then there is a $B=O_1\times O_2\in\mathscr{B}$ such that $x\in B\subseteq W$. Then $\pi_1(B)=O_1\subset X_1$. Since $O_1$ is an element of $\mathscr{B}_{X_1}$, the basis of the factor space $X_1$, then $O_1$ is open. Taking a union of a family of $O_{1,i}$ such that it equals $\pi_1(W)=U_1$, then U_1 is open set. Thus projection mapping to the first coordinate is a quotient map.

Any suggestions?

I sincerely thank you for taking the time to read this question. I greatly appreciate any assistance you may provide.

  • 1
    $\begingroup$ Your answer seems correct, but is a bit messy. It would be clearer if we did not use the $x \in B$ bit, which I believe is unnecessary. I've posted a cleaner method below. $\endgroup$ – Krishan Bhalla Apr 27 '15 at 23:07

If $p: X_1 \times X_2 \rightarrow X_1$ is our projection map, then note for any open $U \subset X_1$, $p^{-1}(U) = U \times X_2$ is an open subset of $X_1 \times X_2$ (why?)

On the other hand, if $U \subset X_1$ is s.t. $p^{-1}(U) = U \times X_2$ is open, then $U$ is open in $X_1$ as open sets in our product topology are unions of elements of our basis, and our basis is $\{ A \times B$ | $A$ open in $X_1, B$ open in $X_2\}$, so in particular, $U$ is the union of open sets in $X_1$, and hence is open itself.


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.