# Are Projection Mappings considered Quotient Maps?

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1. Apr 27, 2015

### Kevin_H

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

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

What I understand:
• Let $X$ be a finite product space and $x=(x_1,x_2,...,x_n)$ a point in $X$. The function $\pi_i:X\rightarrow X_i$ defined by $\pi_i(x)=x_i$ is called the projection map on the $i$th coordinate.
• 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, subset $U$ of $Y$ closed in $Y$ if and only if $f^{−1}(U)$ closed in $X$.
Rough attempt:

To show $\pi_1:X\rightarrow X_1$ is a quotient map, I need to show that (1) $\pi_1$ is surjective, (2) $\pi_1$ is continuous, and (3) for some $U_1\subset X_1, \pi_1^{-1}(U_1)$ open in $X \Rightarrow U_1$ is open in $X_1$.
$\pi_1$ is clearly onto since $\forall x\in X, \exists x_1\in X_1: \pi_1(x)=x_1.$ Let $O_1$ be open subset of $X_1$. Then $\pi_1^{-1}(O_1)=\{x=(x_1,x_2)\in X:x_1\in O_1\}=O_1\times X_2$. Thus $\pi_1^{-1}(O_1)$ is open in $X$. Thus $\pi_1$ is continuous. The third part is what I am not sure of. Since $\pi_1$ is surjective, we have $\pi_1(\pi_1^{-1}(V_1))=V_1, \forall V_1\subset X_1$. Assume now that $W=\pi_1^{-1}(O_1)$ is an open subset of $X$. My question is, how would I conclude that $\pi_1(W)=O_1$ is indeed open. Can I just assume or is there something I am missing? I'm sure I am missing something.

Out of curiosity, could I generalize this and say if $X=\prod_1^nX_i$ if a finite product space, then the first coordinate projection is a quotient map? How about the $i$th coordinate projection? I want so say yes. I can easily prove that $\pi_i:X\rightarrow X_i$ is continuous using the same method I did for continuity above. Anywho, that's probably a different story for a different time.

Sorry for the long read. If I need to clarify on anything I presented, let me know. I sincerely thank you for taking the time to read this post. I greatly appreciate any assistance you may provide.

2. Apr 27, 2015

### Dick

No, you can't just assume it. You want to think about the basis for the product topology. It's all sets of the form $O_1 \times O_2$ where $O_1$ and $O_2$ are open in $X_1$ and $X_2$ respectively. Can you use that to show that if $W$ is an open set in $X_1 \times X_2$ then if $x \in W$ then there is a basis set $O_1 \times O_2$ that contains $x$ and is also a subset of $W$? That would do it, yes?

And sure, there is nothing special about the first coordinate.