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!


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.


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

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 $A$, $$f\left[\bigcup\mathcal{U}\right]=\bigcup_{U\in\mathcal{U}}f[U]\;.$$

  • $\begingroup$ 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

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 $f(U) \subseteq B$.

  • $\begingroup$ 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

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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