# Projection Mappings are Quotient Mappings?

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.

• 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. – Krishan Bhalla Apr 27 '15 at 23:07

## 1 Answer

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.