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I've been going through Munkres' book on topology on my own, and I just struggled through the proof of 10d) from chapter 2 section 19. I've never had a chance to show one of my proofs to anyone, so I suspect my language in general is off-standard, and I'm not sure my proof is right in general. Any verification or criticism would be great! I only included part of my proof, I'm more confident about the rest and it's already fairly long.

Problem: 10d) Let $A$ be a set; let $(X_\alpha)_{\alpha\in J}$ be an indexed family of spaces; and let $(f_\alpha)_{\alpha\in J}$ be an indexed family of functions $f_\alpha: A \rightarrow X_\alpha$. Let $\mathcal{S}_\beta=\{f_\beta^{-1}(U_\beta) | U_\beta \text{ is open in } X_\beta\}$, and let $\mathcal{S}=\bigcup \mathcal{S}_\beta$. $\mathcal{T}$ is the topology on $A$ formed by the subbasis, $\mathcal{S}$. Let $f: A\rightarrow \prod X_\alpha$ be defined by the equation $f(a) = (f_\alpha(a))_{\alpha\in J}$; let $Z$ denote the subspace $f(A)$ of the product space $\prod X_\alpha$. Show that the image under $f$ of each element of $\mathcal{T}$ is an open set of $Z$.

My Proof: Let $U$ belong to the subbasis $\mathcal{S}$ of $\mathcal{T}$, and let $U$ be the preimage of some open set $V\in X_\beta$ for some $\beta$. $f_\beta(U) = f_\beta(f_\beta^{-1}(V)) = V \cap f_\beta(A)$, thus it is an open set in the subspace $f_\beta(A)$.

The set $\prod V_\alpha$ where $V_\alpha = X_\alpha$ for all $\alpha \neq \beta$ and $V_\alpha=V$ when $\alpha=\beta$ is open in $\prod X_\alpha$. The intersection $B = \prod V_\alpha\cap f(A)$ is exactly $f(U)$. Proof proceeds: consider an $x\in U$. $f_\beta(x) \in V$, and for any other $\alpha$, $f_\alpha(x)$ is obviously in $X_\alpha$, so $f(x)\in B$. In the other direction, consider an $x\in B$. $\pi_\beta(x) \in f_\beta(U)$. Suppose there were a $y$ such that $f_\beta(y)=\pi_\beta(x)$, but $f(y)\not\in f(U)$. However$^{note}$, $y\in f^{-1}_\beta(\pi_\beta(x))$, equivalently, $y\in f^{-1}_\beta(f_\beta(U))$, so $y\in U$. Thus, $\pi_\beta(x)$ being an element of $f_\beta(U)$ is a sufficient condition for $x\in U$.

$B$ is clearly an open set of $Z$, so $f(U)$ is an open element of $Z$, where $U$ is an arbitrary subbasis element of $\mathcal{T}$.

Note: where I wrote note as a superscript is just before the part where I think my proof is the most shaky, if there's a major flaw, it's probably in that part.

Thanks for the help!

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Your argument is basically correct as far as it goes, but it doesn’t go far enough, and the last part does need to be cleaned up a little; $y\in f_\beta^{-1}[\{\pi_\beta(x)\}]$ implies that $y\in f_\beta^{-1}[f_\beta[U]]$, but the two statements are not equivalent. Here’s how you could clean up that last part:

You’re starting with an arbitrary $x\in B$, i.e., an $x\in\prod_{\alpha\in J}X_\alpha$ such that $\pi_\beta(x)\in V$ and $x\in f[A]$. Thus, there is as $a\in A$ such that $x=f(a)$. But then

$$\langle\pi_\alpha(x):\alpha\in J\rangle=x=f(a)=\langle f_\alpha(a):\alpha\in J\rangle\;,$$

so $f_\alpha(a)=\pi_\alpha(x)$ for each $\alpha\in J$. In particular, $f_\beta(a)=\pi_\beta(x)\in V$, so $a\in f_\beta^{-1}[V]=U$, and $x=f(a)\in f[U]$. Thus, $B\subseteq f[U]$, as desired.

But as I said, the argument doesn’t go quite far enough. You’ve shown that the image under $f$ of each member of $\mathcal{S}$ is open, but this doesn’t immediately show that $f$ is an open map. Suppose that $U=U_1\cap\ldots\cap U_n$, where each $U_k\in\mathcal{S}$. It’s not necessarily true that

$$f[U]=f[U_1]\cap\ldots\cap f[U_n]\;,$$

so the fact that the sets $f[U_k]$ are open doesn’t ensure that $f[U]$ is open. You can fix this by applying exactly the same ideas to an arbitrary basic open set.

Let

$$U=f_{\alpha_1}^{-1}[V_{\alpha_1}]\cap\ldots\cap f_{\alpha_n}^{-1}[V_{\alpha_n}]$$

be a basic open set, where $\alpha_1,\ldots,\alpha_n$ are distinct members of $J$, and $V_{\alpha_k}$ is open in $X_{\alpha_k}$ for $k=1,\ldots,n$. Let $F=\{\alpha_1,\ldots,\alpha_n\}$, let $V_\alpha=X_\alpha$ for $\alpha\in J\setminus F$, and let $V=\prod_{\alpha\in J}V_\alpha$; you want to show that $f[U]=V\cap f[A]$.

  • See if you can adapt your proof that $f[U]\subseteq V\cap f[A]$ when $U\in\mathcal{S}$ to handle this slightly more general case.
  • Then see if you can similarly adapt my argument above for $V\cap f[A]\subseteq f[U]$ when $U\in\mathcal{S}$.

Both adaptations are pretty straightforward, but feel free to leave a question if you get stuck.

Note that once you’re proved that $f$ sense basic (instead of just subbasic) open sets to open sets, you can conclude that the map $f$ is open, because if $\mathcal{U}$ is any family of subsets of $A$, $$f\left[\bigcup\mathcal{U}\right]=\bigcup_{U\in\mathcal{U}}f[U]\;.$$

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  • $\begingroup$ Right, I did extend my proof to basis elements, I just decided to only post this part because this was what I was least sure about, and it was already long enough that I figured it would be best not to add more. $\endgroup$ – drowdemon Aug 22 '15 at 3:40
  • $\begingroup$ @drowdemon: In that case you were basically okay, just needing a little more care in the writing. $\endgroup$ – Brian M. Scott Aug 22 '15 at 3:43
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My first thought on reading this was Wow! they're still using Munkres? But then I pulled out my copy, and chapter 2 only has 11 sections, so I guess it's been updated a little...

Unfortunately $y \in f^{-1}(f(U))$ does not imply that $y \in U$. For example $f(x) = x^2, U = \{1\}$ and $y = -1$. (Edit: while true, this is not applicable here. See comments.)

More generally, I believe you are wrong about $B = f(U)$. All you can show (which you have shown) is that $f(U) \subseteq B$.

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  • $\begingroup$ I know that generally that's not true, but in this case $U = f^{-1}_\beta(V)$, and $f_\beta(U) = V\cap f_\beta(A)$, and so $f^{-1}(V\cap f_\beta(A))\in U$, so $y\in U$. I don't see any flaws in that, but I do think the whole thing is kinda shakey and my assertion that $B=f(U)$ does feel like it might be too bold. And maybe they don't use Munkres, that just seemed to be a popular book I found online :P They probably just renumbered the sections, chapter 2 only has 11, but the numbering continues from chapter 1, so it starts at chapter 2 section 12. $\endgroup$ – drowdemon Aug 21 '15 at 15:31
  • $\begingroup$ The OP is correct about $B=f[U]$, and the argument given is basically correct. There is a real problem with the argument as a whole, but it lies elsewhere: in order to show that the map $f$ is open, it’s not sufficient to show that it takes subbasic open sets to open sets. See my answer. $\endgroup$ – Brian M. Scott Aug 21 '15 at 23:23
  • $\begingroup$ @drowdemon - You are correct. $\endgroup$ – Paul Sinclair Aug 21 '15 at 23:26
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Exercise 10, section 19, Chapter 2 (Munkres).

Let $A$ be a set; let $\{ X_\alpha \}_{\alpha \in J}$ be an indexed family of spaces; and let $\{ f_\alpha \}_{\alpha \in J}$ be an indexed family of functions $f_\alpha : A \to X_\alpha$.

  • (a) Show there is a unique coarsest topology $\mathcal{T}$ on $A$ relative to which each of the functions $f_\alpha$ is continuous.

  • (b) Let

$$\mathcal{S}_\beta = \{ f_\beta^{-1} \left ( U_\beta \right ) | U_\beta \mbox{ is open in } X_\beta \},$$

and let $\mathcal{S} = \bigcup \mathcal{S}_\beta$. Show that $\mathcal{S}$ is a subbasis for $\mathcal{T}$.

  • (c) Show that a map $g : Y \to A$ is continuous relative to $\mathcal{T}$ if and only if each map $f_\alpha \circ g$ is continuous.

  • (d) Let $f : A \to \prod X_\alpha$ be defined by the equation

$$f \left ( a \right ) = \left ( f_\alpha \left ( a \right ) \right )_{\alpha \in J};$$

let $Z$ denote the subespace $f(A)$ of the product space $\prod X_\alpha$. Show that the image under $f$ of each element of $\mathcal{T}$ is an open set of $Z$.

Proof of part (d).

  • Step 1. Let $B$ be a basis element for the topology $\mathcal{T}$ on $A$. Then $B$ is a finite intersection of elements of the subbasis $\mathcal{S}$. Without lost of generality, suppose that $B$ is the intersection of the sets $V_{\beta_i} = f_{\beta_i}^{-1}( U_{\beta_i} ) \in \mathcal{S}$, where $U_{\beta_i}$ is an open set of the space $X_{\beta_i}$ for each $\beta_i \; (i = 1, \ldots, k)$.

Let $\prod U_\alpha$ be a subset of $\prod X_\alpha$, where $U_\alpha = X_\alpha$, except for a finite number of indices for which $U_\alpha = U_{\beta_i} \; (i = 1, \ldots, k)$. Then, by definition, $\prod U_\alpha$ is a basis element for the product topology on $\prod X_\alpha$.

  • Step 2. Let $Z$ be the subespace $f(A)$ of the product space $\prod X_\alpha$, and let

$$\mathcal{T}_Z = \left \{ Z \cap U \, | \, U \, \mbox{abierto en} \, \displaystyle\prod X_\alpha \right \}$$

be the subespace topology on $f(A)$.

If $x$ is a point of $B$, and $\alpha = \beta_i \; (i = 1, \ldots, k)$, then $x$ belongs to one of the sets $V_{\beta_i} = f_{\beta_i}^{-1}( U_{\beta_i} )$ of the subbasis $\mathcal{S}$. Hence, the image $f_{\beta_i}( x )$ belongs to the open set $U_{\beta_i} \subset X_{\beta_i} \; (i = 1, \ldots, k)$. And if $\alpha \ne \beta_i \; (i = 1, \ldots, k)$, the image $f_\alpha( x )$ belongs to $U_\alpha = X_\alpha$. Therefore, the image ${\bf y} = f ( x ) = (f_\alpha(x))_{\alpha \in J}$ belongs to the product $\prod U_\alpha$ defined in Step 1. On the other hand, ${\bf y} = f ( x )$ also belongs to $f(A) \subset \prod X_\alpha$. That is ${\bf y} = f ( x )$ is a point of the intersection of $\prod U_\alpha$ with $f(A)$, and then it is clear that $f(B)$ is an open set of $\mathcal{T}_Z$.

  • Step 3. Now, let $W$ be an open set of the topology $\mathcal{T}$ on $A$. Then, $W$ is an union of basis elements for the topology $\mathcal{T}$. And its image $f(W)$ is an union of the images under $f$ of the basis elements whose union is $W$. Since these images belong to $\mathcal{T}_Z$, by Step 2, it follows that $f(W)$ is also a member of $\mathcal{T}_Z$.
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