Let $I$ be an index set, $X_i$ be a topological space for each $i \in I$ and $X = \prod_{i \in I} X_i$ the product of all $X_i$. Then the product topology is exactly the initial topology with respect to the canonical projections $\pi_i: X \to X_i$.

Let $A$ be a topological subspace of $X$ (i.e. $A$ has the subspace topology in $X$) and $\varphi_i$ are the restriciton of the projections $\pi_i$ on $A$.
Question: Is $A$ the initial topology with respect to the maps $\varphi_i$?

I think that the arguments which I need can be found in this nice explanation of Henno Brandsma, but I am unable to convert it into a line of reasoning which convinces me. Could you please answer my question and give an explanation? Thank you!


It's a special case of the transitive law of initial topologies, which is shown in my linked answer:

$A$ has the initial topology wrt $i_A: A \to X$.

$X$ has the initial topology wrt the maps $\pi_i: X \to X_i, i \in I$.

It follows that $A$ has the initial topology wrt the maps $\pi_i \circ i_A: A \to X_i, i \in I$, and $\pi_i \circ i_A$ is exactly $\varphi_i=\pi_i|_A$.

In that answer the original question was whether on $A=\prod_i A_i$ the product topology of subspace topologies coincides with the subspace topology of the large product, but the same law answers both questions.

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    $\begingroup$ So the answer is yes when the subspace is a product of subspaces. What if the subspace isn't a product of subsoaces, the closed unit disk for example? $\endgroup$ – William Elliot Feb 9 '19 at 13:27
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    $\begingroup$ @WilliamElliot It's true in both cases. In my answer $A$ is any subspace of the product. $\endgroup$ – Henno Brandsma Feb 9 '19 at 13:27
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    $\begingroup$ @mrtaurho The last edit is correcting the ording of composition. If you take a look at the domains and codomains, you'll know that the original answer contains a small mistake. Since such small typo would alter the mathematical meaning greatly, it's OK to change them. Any 2k user can rollback any edits. Btw, you should have used "conflic with author's intent" in this review. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 9 '19 at 16:11
  • $\begingroup$ @Diglett the edit is fine. $\endgroup$ – Henno Brandsma Feb 9 '19 at 16:39

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