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In Aluffi's book "Algebra. Chapter 0" a group homomorphism is defined the following way. For two sets $G$ and $H$ with operations $m_G$ and $m_H$ firstly a set-function $\varphi\colon G\to H$ is considered. Then the author states:

Note that the set-function $\varphi\colon G\to H$ determines a function $$ (\varphi\times\varphi)\colon G\times G\to H\times H: $$ we could invoke the universal property of products to obtain this function.

He then defines $\varphi\times\varphi$ explicitly as $$ (\varphi\times\varphi)(a,b)=(\varphi(a),\varphi(b)) $$ and requires a certain diagram to commute.

But somehow I fail to understand how exactly $\varphi\times\varphi$ is a product in this case. According to the universal property of products it is a final object in category $\mathsf{C}_{A,B}$ ($\mathsf{C}=\mathsf{Set}$, and $A=B=H$, I assume). So I took $H\times H$ with two morphisms $\tau_1,\tau_2\colon H\times H\to H$ (natural projections) and tried to construct a morphism from another alike object $G\times G$ (which also has natural projections $\pi_1,\pi_2)$, namely $\sigma\colon G\times G\to H\times H$. What I got indeed strongly resembled a product, not $\varphi\times\varphi$ though, but rather $\varphi\pi_1\times\varphi\pi_2$. The diagram I got commutes if $\varphi\pi_1=\tau_1\sigma$ and $\varphi\pi_2=\tau_2\sigma$.

In general, according to the definition, $f_A\times f_B$, $f_A$ and $f_B$ all have the same domain (for example, $f_A\times f_B\colon Z\to A\times B$, $f_A\colon Z\to A$, $f_B\colon Z\to B$), which makes me even more confused.

What do I miss here?

AndrewNova
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  • $\varphi \times \varphi$ denotes the action of the product functor $- \times - : \mathrm{Set} \times \mathrm{Set} \to \mathrm{Set}$ on the pair of morphisms $(\varphi, \varphi)$, which is a morphism in $\mathrm{Set} \times \mathrm{Set}$ between $(G, G)$ and $(H, H)$. – Naïm Camille Favier Nov 27 '24 at 15:04

2 Answers2

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Here Aluffi is invoking the universal property of products for $H \times H$, not stating that $\varphi \times \varphi$ is a product. This former property states that to give a map $G \times G \to H \times H$ it is enough to give two functions $f : G \times G \to H$ and $g : G \times G \to H$. Can you see what $f$ and $g$ are here?

Fernando Chu
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What you've been writing as $f \times g$ (the morphism into a product $Z \rightarrow A \times B$ induced by $f: Z \rightarrow A$ and $g: Z \rightarrow B$) is usually written as $(f,g)$. I'll use this notation for the rest of this answer, since it frees up the $f \times g$ notation to be used for something else.

To be specific, given $f:A \rightarrow B$ and $g: C \rightarrow D$ we can use $f \times g: A \times C \rightarrow B \times D$ to mean $(f \pi_1 , g \pi_2)$. This is the morphism that, as you noticed, satisfies $\tau_1(f \times g) = f \pi_1$ and $\tau_2(f \times g) = g \pi_2$.

The important thing about this $f \times g$ construction is that it makes the product into a functor $\mathscr{C} \times \mathscr{C} \rightarrow \mathscr{C}$. You should check to see how this works yourself.

Chessanator
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  • Thank you for your answer! I wrote $\varphi\times \varphi$ only because that's how it is written in the book. It happened to match a morphism into a product notation, when products were defined. So, is thus $\varphi\times \varphi$ not a product? Is all this a purely notational issue then? – AndrewNova Nov 27 '24 at 15:27
  • @AndrewNova You are looking at two different things. If you have maps from a single set $Z$ to $A$ and to $B$, $f\colon Z\to A$ and $g\colon Z\to B$, then you get a unique map from $Z$ into $A\times B$, which is written $(f,g)\colon Z\to A\times B$. The other notation is when you have two maps from potentially different sets, $f\colon Z\to A$, and $g\colon W\to B$, and then these maps give you a map from $Z\times W$ to $A\times B$. This map is called $f\times g$. – Arturo Magidin Nov 27 '24 at 17:55
  • @ArturoMagidin, thank you for your reply! I do realize that these two particular notions are different. I think, that the main thing which confused me was that Aluffi in I.5.4. Products, explaining products, used the notation $f_A\times f_B$ to denote a morphism from some $Z$ into $A\times B$ ($f_A\colon Z\to A$, $f_B\colon Z\to B$). Now I just don't understand why did he use this specific notation there. – AndrewNova Nov 27 '24 at 18:08
  • I have found this answer referring to basically the same question. It turns out both notations are possible, and it is a matter of context most of the times. I appreciate your answers and the suggestion to use $(f,g)$ to denote a morphism into a direct product of objects. – AndrewNova Nov 27 '24 at 19:25