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If $Y$ and $X_i$ are topological spaces and $\prod_{i\in I}X_i$ has the product topology, how can I test continuity of the map $f:Y\rightarrow \prod_{i\in I}X_i$ using the projections?

This question is related to a problem I posted here: $f: \mathbb{R}\rightarrow \mathbb{R}^{\mathbb{R}}: x\mapsto (e^{t\sin(x)})_{ t\in \mathbb{R}}$

k.stm has solved this using the product topology. My problem is that I cannot see the connection between projections and this proof.

I'm posting this here as a general question because I want to investigate the general approach to such problems.

Averroes2
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    The map $f$ is continuous if and only if the compositions $p_i \circ f \in \mathrm{Hom}_{\mathbf{Top}}((Y, \mathscr{S}), (X_i, \mathscr{T}_i))$ are continuous for every index $i \in I$ (the cursive letters denote the topologies on each of the spaces in question). This follows straight from the definition of the product topology. – ΑΘΩ Sep 24 '20 at 07:48
  • @ΑΘΩ Why not an official answer? – Paul Frost Sep 24 '20 at 10:30
  • @Paul Frost Sir, I appreciate your encouragement. I thought the matter to be such an absolutely elementary statement that it would not be deserving of official answer status. Seeing however that I enjoy your endorsement, may I proceed to render it into a full-fledged answer (brief as it is). – ΑΘΩ Sep 24 '20 at 10:42
  • @ΑΘΩ You are of course right, it is elementary. However, if nobody gives an answer the question will remain forever in the "uanswered" queue and attract attention of readers although nothing is open. – Paul Frost Sep 24 '20 at 10:46
  • I have used what is known as the universal property for products. – k.stm Sep 24 '20 at 14:58

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Given a family $X$ of sets indexed by $I$ and a family $\mathscr{T} \in \displaystyle\prod_{i \in I}\mathscr{Top}(X_i)$ of topologies on each of the sets $X_i$, let us denote by $\displaystyle\bigotimes_{i \in I}\mathscr{T}_i$ the direct product topology on the cartesian product $\displaystyle\prod_{i \in I}X_i$. If $p_i \colon \displaystyle\prod_{i \in I}X_i \to X_i$ denotes the canonical projection, then it follows straight from the definition that given an arbitrary topological space $(Y, \mathscr{S})$ and a map $f \colon Y \to \displaystyle\prod_{i \in I}X_i$ we have the equivalence: $$f \in \mathrm{Hom}_{\mathbf{Top}}\left((Y, \mathscr{S}), \left(\displaystyle\prod_{i \in I}X_i, \displaystyle\bigotimes_{i \in I}\mathscr{T}_i\right)\right) \Leftrightarrow (\forall i)\left(i \in I \Rightarrow p_i \circ f \in \mathrm{Hom}_{\mathbf{Top}}((Y, \mathscr{S}), (X_i, \mathscr{T}_i))\right),$$ which is a very stiff and formal way of saying that $f$ is continuous if and only if all the compositions $p_i \circ f$ are continuous for every $i \in I$.

In the terminology Bourbaki is an adept of, the direct product topology is the initial structure induced by the family of spaces $\left(X_i, \mathscr{T}_i\right)_{i \in I}$ on the cartesian product $\displaystyle\prod_{i \in I}X_i$ via the family of canonical projections.

ΑΘΩ
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  • As to the last remark : I have a rather long post on initial topologies in this answer where I also derive this criterion in greater generality. We did this stuff in our first course of general topology at the time. – Henno Brandsma Sep 24 '20 at 21:50
  • @Henno Brandsma You had an admirable degree of patience to write it all down. Perhaps the OP would also profit from this knowledge, so may I suggest that you also let him know of your detailed posting. – ΑΘΩ Sep 25 '20 at 01:16
  • @Henno Brandsma: thank you. This helps – Averroes2 Sep 25 '20 at 05:16
  • @Averroes2 you’re welcome. It shows how common this idea is. Looking at it abstractly allows some individual facts to be unified under one heading. – Henno Brandsma Sep 25 '20 at 05:22
  • @Henno Brandsma Sir, if you permit me to take advantage of your expertise in topology, I might want to ask you a question: in your comprehensive posting you mention "$k$-spaces" and in connection to them make the assertion that any first countable $T_1$ space is in particular a $k$-space. Why would this be so? From what I know, any first countable space can be shown to be countably generated where by this syntagm I mean the equivalence "closed subset iff any intersection with a compact subset is relatively closed in that subset" (and compactness in this acceptation does not include $T_2$). – ΑΘΩ Sep 25 '20 at 10:41
  • @ΑΘΩ if $A \subseteq X$ is not closed we have a sequence $(a_n)$ in $A$ and some $ p \notin K$ to which this sequence converges. This by first countability. Then $K={a_n\mid n \in \Bbb N}\cup {p}$ is compact and $A\cap K $ is not closed in $K$. So $X$ is a $k$-space. – Henno Brandsma Sep 25 '20 at 17:56
  • @Henno Brandsma Thank you for the clarification. It turns out that we were speaking about the same result (for a moment I had the impression that the result you were mentioning in your detailed posting was slightly different). There is this degree of variability in the definition of compactly generated/$k$-spaces that can at times be a tad confusing. – ΑΘΩ Sep 26 '20 at 02:11