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This seems very obvious, and that may be the reason why I cannot find this online. I think it's yes, and my proof is that the isomorphism between the products is the product of the isomorphisms: Given

$$\varphi_1: A \to B,\ \varphi_2: C \to D,$$

define $$\varphi(a \times c) = \varphi_1(a) \times \varphi_2(c)$$

and then show that $\varphi$ is injective, surjective and a homomorphism, all of which follow from injectivity, surjectivity and homomorphism property of $\varphi_1$ and $\varphi_2$.

First question, am I correct?

Second question, what's the relation between the above, this exercise from Artin Algebra, this exercise from Munkres Topology and this theorem from Munkres Topology?

  • Exercise 2.11.8 in Artin Algebra

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  • Exercise 18.10 in Munkres Topology

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  • Theorem 18.4 ("Maps into products") in Munkres Topology:

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By the way, the context for my questions is this theorem in Munkres Topology:

  • Theorem 67.8 in Munkres Topology

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    $\begingroup$ Do You mean in the title "Show that $A\times C$ and $B\times D$ are isomorphic"?? $\endgroup$ Nov 8 '18 at 12:40
  • $\begingroup$ @PeterMelech Thank you! $\endgroup$
    – user198044
    Nov 8 '18 at 12:42
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    $\begingroup$ You are welcome! First question: Yes You are, just use the group structure as it is defined for the products and you will succeed. $\endgroup$ Nov 8 '18 at 12:47
  • $\begingroup$ @PeterMelech Thank you! $\endgroup$
    – user198044
    Nov 8 '18 at 12:51
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Yes, this is true.

In general a product in a category preserves morphisms, (and so does coproduct).

In other words if $\mathcal{C}$ is a category with a product $(\times)$ and coproduct $(\ast)$, and $A,B,C,D \in \operatorname{obj}\mathcal{C}$, such that there exist morphisms $\phi : A \to B$ and $\varphi : C \to D$, then there exist unique morphisms corresponding to product of our initial morphisms: $$(\phi \times \varphi) : A \times C \to B \times D$$ And a unique morphism corresponding to coproduct of our initial morphisms: $$(\phi \ast \varphi) : A \ast C \to B \ast D$$ If both of our morphisms $\phi, \varphi$ are isomorphisms, then both $(\phi \times \varphi)$ and $(\phi \ast \varphi)$ are isomorphisms as well. All of this is due to properties of product and coproduct, this is true from universal property. This is the connection between all those books I suppose, they just work in different categories:

  • ${\bf Top}$ - topological spaces, where morphisms are continuous maps and isomorphisms are homeomorphisms (conitnuous map with continuous inverse); and products and coproducts are given by Cartesian product and wedge.
  • ${\bf Grp}$ - groups, where morphisms are homomorphisms and isomorphisms are, well... isomorphisms; and products and corproducts are given by direct product and free product.
  • ${\bf Ab}$ - Abelian groups) which are subcategory of groups, in which, interestingly enough, products and corproducts coincide, (free product is the same as direct product).

Not sure if this is what you're looking for though.

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  • $\begingroup$ Thank you! So the relationship is that these follow from "categories" in general, with the way you just described? I haven't learned any category theory yet. $\endgroup$
    – user198044
    Nov 9 '18 at 12:54
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    $\begingroup$ You can prove those properties for respective operators and morphisms in each respective category, but that's how it all interconnects in general. There are special maps called functors, which map morphisms to each other. Some nicer functors preserve the relations given by products and coproducts, for example a common example is a fundamental group of a topological space, which can be thought of as a functor $\pi_1 : {\bf Top} \to {\bf Grp}$, a fundamental group of a cartesian product of topological spaces is a direct product of their respective fundamental groups, and same for coproducts. $\endgroup$ Nov 9 '18 at 22:53
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    $\begingroup$ The fundamental group is also "weaker", i.e. the homeomorphism between two topological spaces implies isomorphism between their fundamental groups, but converse is not necessarily true, and requires a different, weaker condition. $\endgroup$ Nov 9 '18 at 22:57
  • $\begingroup$ Eetu Koskela, I don't understand anything after your first sentence, hehe. Thank you! $\endgroup$
    – user198044
    Nov 10 '18 at 3:15

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