0
$\begingroup$

Let $(Y,\tau)$ and $(X_i,\tau_i)$, $i= 1,2,\ldots,n$ be topological spaces. Further for each $i$, let $f_i$ be a mapping of $(Y,\tau)$ into $(X_i, \tau_i)$. Prove that the mapping $$f:(Y,\tau)\longrightarrow \prod(X_i,\tau_i),$$ given by $f(y) =\langle f_1(y),f_2(y),\ldots,f_n(y)\rangle$,is continuous if and only if every $f_i$ is continuous. Hint: $f_i=p_i(f)$ where $p_i$ is the projection mapping of the $i$th component.

This seems trivial for both directions so I'm not sure if I'm missing something.

Assume each $f_i$ is continuous. The take an open set $A$ in the product topology where its basis are open sets from each $X_i$. Since $f$ is continuous, the preimage of $A$ in the product topology is an $n$-tuple of open sets in $Y$ which means each component $f_i$ is continuous.

For the other direction, we assume $f$ is open and show each $f_i$ is open. Using the hint, each $f_i=p_i(f)$ which is a composition of two continuous functions since the projection mapping is continuous. Then each $f_i$ has to be continuous.

$\endgroup$
3
  • $\begingroup$ It is as easy as that. You have a correct proof. $\endgroup$ Feb 1 '21 at 12:31
  • $\begingroup$ Use MathJax please, see here, e.g. $\endgroup$ Feb 1 '21 at 13:07
  • $\begingroup$ The $f$ open part is not irrelevant, is it? The quoted question asks for continuity, not openness. $\endgroup$ Feb 1 '21 at 13:08
1
$\begingroup$

Your proof is almost correct. If $f = (f_1, \ldots, f_n)$ and all $f_i$ are continuous, then for a basic open subset $O=\prod_{i=1}^n O_i$, where $O_i \subseteq X_i$ is open for all $i$, we have that $$f^{-1}[O] = \{y \in Y\mid f(y) \in \prod_{i=1}^n O_i\}= \{y \in Y\mid \forall i \in \{1, \ldots n\}: f_i(y) \in O_i \} = \bigcap_{i=1}^n f_i^{-1}[O_i]$$

which is a finite intersection (not product or $n$ tuple!) of open subsets by the continuity of the component mappings $f_i$, and hence open. As inverse images under $f$ of basic open subsets are open, the same holds for all open subsets of the product topology and so $f$ is continuous.

The reverse indeed simply follows from $f_i = \pi_i \circ f$ for all $i$ and the fact that compositions of continuous maps are continuous.

It's easy to see how this generalises to infinite products (if you've treated these as well), but not to the box topology on such a product.

$\endgroup$
5
  • $\begingroup$ Thanks. By the way, just a question about most of these proofs in product topology. Open sets in a product topology are not just basic open sets (product of open sets of each component topology which only forms a basis but not the whole product topology), but also all their unions. A lot of proofs assumes we are only dealing with basic open sets which are just the product of open sets of each topology. Is that always sufficient or allowed? In this specific case, since the preimage of unions are unions of preimage, it seems ok. $\endgroup$
    – willyx888
    Feb 2 '21 at 0:03
  • $\begingroup$ oh one question about the intersection for the first part of the proof, where did that come from? Are we assuming each fi is like the projection mapping where the preimage of say pi(U) = X1 x X2 x X3 x X4 x pi^-1(U) x X5.....Xn? I'm not sure if that is the case for each fi though. Each fi maps Y to Xi, so the preimage of each fi is just elements of Y instead of a n- tuple so there should be no need to take the intersection at the end right? $\endgroup$
    – willyx888
    Feb 2 '21 at 2:15
  • $\begingroup$ @willyx888 Yes, basic elements are enough in continuity proofs, for the reason you mentioned. Often this is some separate proposition or theorem. It's one of the advantages of working with a base of more concrete sets instead of having to work with all of them. You do need the intersection because you need the simultaneous condition that $f_i(y) \in X_i$ for all $i$ in order for $f(y)$ to be in the rectangle set $U_1 \times \ldots \times U_n$. But you wrote in the question that the inverse image consisted of $n$-tuples which is not true, it's a subset of $Y$. $\endgroup$ Feb 2 '21 at 6:46
  • $\begingroup$ Ahh that makes senses. Yes the inverse image are all just subsets of Y. It's like taking the minimum of finite sets in real analysis proofs. $\endgroup$
    – willyx888
    Feb 2 '21 at 9:00
  • $\begingroup$ @willyx888 Indeed, you could see it a bit like that. It's like why finite intersections of open sets are open but not necessarily all infinite ones. $\endgroup$ Feb 2 '21 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.