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I think this is a silly question, but I'm having troubles with the abstraction in category theory. Let $\mathcal{C}$ be a category and $A,B$ two objects in $\mathcal{C}$. Prove that if $A \times B$ exists, then $B \times A$ exists.

Now, $A \times B$ exists means that $A \times B$ is an object of $\mathcal{C}$ and there are two arrows $\pi_1 :A \times B \rightarrow A$, $\pi_2: A \times B \rightarrow B$ such that for any object $S$ in $\mathcal{C}$ and for any arrows $f_1: S \rightarrow A$, $f_2: S \rightarrow B$ there exists a unique arrow $u: S \rightarrow A \times B$ such that $f_1 = \pi_1 \circ u$, $f_2 = \pi_2 \circ u$.

Let $S'$ be an object of $\mathcal{C}$ and $f_1': S' \rightarrow A$, $f_2': S' \rightarrow B$. In order to prove that $B \times A$ exists I have to prove that this is an object of $\mathcal{C}$ and there are two arrows $p_1: B \times A \rightarrow B$, $p_2: B \times A \rightarrow A$ such that there's a unique $u': S' \rightarrow B \times A$ with $f_1'=p_2 \circ u'$, $f_2'=p_1 \circ u'$. But for that $(S',f_1',f_2')$ I know that $f_1'= \pi_1 \circ u$. And now I'm stuck.

Any help please? Thanks

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    $\begingroup$ The only object you know to exist (apart from $A$ and $B$) is $A\times B$ -- so what would you try to put in charge for the role of $B\times A$? $\endgroup$ Jul 6 at 14:21
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It may help if you avoid the names $A \times B$ and $B \times A$ of the objects that 'exist'.

You know there is some object $P$ with two arrows $\pi_1 \colon P \to A$ and $\pi_2 \colon P \to B$ satisfying the universal property of the product of $A$ and $B$. You have to find some object $Q$ with two arrows $p_1 \colon Q \to B$ and $p_2 \colon Q \to A$ satisfying the universal property of the product of $B$ and $A$. So, you just take $Q = P$, $p_1 = \pi_2$ and $p_2 = \pi_1$ and everything works out just fine.

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    $\begingroup$ $p_1$ should have domain $Q$, but otherwise the most direct and best explanation. $+1$ $\endgroup$
    – Randall
    Jul 6 at 14:35
  • $\begingroup$ @Randall Indeed; I fixed it. Thanks. $\endgroup$ Jul 6 at 14:35
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    $\begingroup$ But, is this answer the proof that $A \times B = B \times A$? I don't think that's true in general... $\endgroup$
    – Ejrionm
    Jul 6 at 18:07
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    $\begingroup$ @Ejrionm No, that's not what this proves. What it does say if $(P,\pi_1,\pi_2)$ is the categorical product of $A$ and $B$, then $(P,\pi_2,\pi_1)$ is the categorical product of $B$ and $A$. $\endgroup$ Jul 6 at 18:46
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    $\begingroup$ @Ejrionm you can NEVER prove equality because the product is only unique up to isomorphism, anyway. $\endgroup$
    – Randall
    Jul 6 at 23:41

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