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Im working through Bloch's Proofs and Fundamentals and exercise 4.3.11 is

Let $B$ be a set, let $A_i,\cdots,A_\kappa$ be sets for some $k\in\mathbb{N}$ be a subset for all $i\in\left\{1,\cdots,\kappa\right\}$ and let $f:B\rightarrow A_1\times\cdots\times A_\kappa$ be a function. Prove that \begin{align} f^{-1}\left(U_1\times\cdots\times U_\kappa\right)=\bigcap_{i=1}^\kappa \left(f_i\right)^{-1}\left(U_i\right), \end{align} where $f_i$ are the coordinate functions of $f$.

My (perhaps poor) attempt at a short proof:

proof Suppose $b=f^{-1}\left(U_1\times\cdots\times U_\kappa\right)$ for some $b\in B$. Then \begin{align}f\left(b\right)&=f\left(f^{-1}\left(U_1\times\cdots\times U_\kappa\right)\right)\\&=U_1\times\cdots\times U_\kappa\subseteq A_1\times\cdots\times A_\kappa.\end{align} Now suppose $b=\bigcap_{i=1}^\kappa\left(f_i\right)^{-1}\left(U_i\right)$ for some $b\in B$. Then \begin{align}f_i\left(b\right)&=f_i\left(\bigcap_{i=1}^\kappa\left(f_i\right)^{-1}\left(U_i\right)\right)\\&=\bigcap_{i=1}^\kappa f_i\left(\left(f_i\right)^{-1}\left(U_i\right)\right)\\&=\bigcap_{i=1}^\kappa U_i\\ &=?\end{align}

I'm not sure if I'm heading in the right direction or if it's wrong altogether. Constructive criticism and hints would be great.

Andrés E. Caicedo
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bjd2385
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    Your notation is a bit off. $f^{-1}(U_1\times\dots\times U_k)$ is a subset of $B$, not an element of $B$. Other than that, $f(f^{-1}(U_1\times\dots\times U_k))\subset U_1\times \dots\times U_k$, but they are not quite equal. I can write out something a bit more substantial if you want. – Moya May 17 '15 at 01:05
  • @Moya Thanks! Does it appear as though I am heading in the right direction though? I feel off-track in the latter portion of the proof, i.e. considering the other side. And yes, I will make those necessary corrections. – bjd2385 May 17 '15 at 01:09
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    You can prove this for $k=2$ and then use induction on $k$ – L F May 17 '15 at 01:19

1 Answers1

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Let $X=f^{-1}(U_1\times \dots \times U_k)$ and let $Y=\bigcap_{i=1}^k f_i^{-1}(U_i)$. We will show that $X\subset Y$ and $Y\subset X$.

Let $x\in X$. It means $f(x)\in f(X)\subset U_1\times \dots\times U_k$. Denoting $f(x)=(f_1(x),\dots, f_k(x)), f_i(x)\in A_i$, it means $f_i(x)\in U_i$ for every $i$, implying $x\in f_i^{-1}(U_i)$ for every $i$. So $x$ is in their intersection, meaning $x\in Y$. So $X\subset Y$.

Conversely, let $y\in Y$. Then $y\in f_i^{-1}(U_i)$ for every $i$. This implies $f_i(y)\in U_i$ for every $i$, so $f(y)=(f_1(y),\dots, f_k(y))\in U_1\times \dots \times U_k$. This means $y\in f^{-1}(U_1\times\dots U_k)=X$, so $Y\subset X$ and they're equal.

Moya
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  • Thank you so much for your answer! I don't think I stated it in my original question, but this was not a homework problem, I'm just studying on my own time some more intricate concepts that weren't covered in my Discrete Math class (I'm an undergrad). I've worked through your solution step by step and it makes total sense. – bjd2385 May 17 '15 at 02:00
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    No worries. The basic idea to these naive set theory relations is just breaking down the definitions of both sides and making sure they line up. In this case, the definition of the intersection/product of sets as well as inverse images. – Moya May 17 '15 at 02:03
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    Also, Munkres' Topology has a good couple of chapters (the first 10 roughly) on this stuff with more challenging types of these sort of problems. – Moya May 17 '15 at 02:06
  • Thanks! I just looked it up and found a copy. I will look it over tonight and the following few weeks, even try many of the problems if I can. – bjd2385 May 17 '15 at 02:09