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I'm having a pretty hard time with this. I'm asked to show that, in the category of sets with exactly 17 elements, no two objects have either a direct product nor direct sum. Part of me doesn't even believe this statement—but whenever I try to come up with a direct product, I get snagged.

Let $(C, \alpha, \beta)$ be a [potential] direct product of $A$ and $B$. Fix some object, $C'$, with mappings, $\alpha'$ and $\beta'$, from $C'$ to $A$ and $B$ respectively. We need a unique $\gamma: C' \rightarrow C$ such that $\alpha \circ \gamma = \alpha'$ and $\beta \circ \gamma = \beta'$.

  • $C = A \times B$ just can't work, because $A \times B$ necessarily has more than 17 elements (in fact, no Cartesian-product-like $C$ can work, because the number of elements is fixed). What about some 17-element subset of $A \times B$? (In the category of sets, $\alpha$ and $\beta$ are injective, but not necessarily surjective). But, what if $\alpha'$ and $\beta'$ are both surjective? So, that can't work, because there's no $\gamma$ that could satisfy this (doesn't that mean that, in the general category of sets, $\alpha$ and $\beta$ have to also be surjective? If they don't "touch" every element in both $A$ and $B$, then one can just define a $\alpha'$ or $\beta'$ that touches the elements $\alpha$ or $\beta$ don't—thus making impossible a direct product.)

  • Let $\alpha(c_n) = a_n$ and $\beta(c_n) = b_n$. This contains bijective $\alpha$ and $\beta$, but all we have to do is define a $C'$ such that $\alpha'(c_1) = a_1$ and $\beta'(c_1) = b_2$.

Ok. So, $\alpha$ and $\beta$ have to be bijective. Let's try to prove this via negation: Since these mappings are necessarily bijective, they have to have an inverse. Thus, $\gamma$ must be such that $\gamma = \alpha^{-1} \circ \alpha'$ and $\gamma = \beta^{-1} \circ \beta'$. To show that we can choose an object $C'$ where $\gamma$ can't make the graph commute, just choose $\alpha'$ and $\beta'$ such that $\alpha^{-1} \circ \alpha \neq \beta^{-1} \circ \beta$.

  1. Can I assume that such an $\alpha'$ and $\beta'$ will always exist?
  2. Was there no point to the number of elements being $17$ specifically? This all seems to work for any category of sets with a fixed number of elements.
  3. Is there something crucial I'm missing?
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  • $\begingroup$ What are the morphisms of your category? $\endgroup$ – J.-E. Pin Jul 21 '15 at 16:53
  • $\begingroup$ Just set-mappings; the same as the morphisms in the more general category of sets. $\endgroup$ – AmagicalFishy Jul 21 '15 at 17:01
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    $\begingroup$ The product projections in $\mathsf{Set}$ don't have to be epic. Consider $\emptyset \times X \to X$. $\endgroup$ – Martin Brandenburg Jul 22 '15 at 9:21
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    $\begingroup$ Notice $\emptyset \times X = \emptyset$. The unique map $\emptyset \to X$ is surjective if and only if $X=\emptyset$. This follows from the definitions of "surjective", "map", "empty set". It is not a convention. $\endgroup$ – Martin Brandenburg Jul 23 '15 at 13:02
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    $\begingroup$ In set theory, $A \times B$ is a set of ordered pairs. It itsself is not an ordered pair. If $A$ is empty, or $B$ is empty, then $A \times B$ is empty, too. $\endgroup$ – Martin Brandenburg Jul 23 '15 at 14:19
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Let $n \geq 2$. If $A,B$ have a product $P$ in the category of sets with $n$ elements, then $\hom(A,P) \cong \hom(A,A) \times \hom(A,B)$ shows $n^n = n^n \cdot n^n$, a contradiction.

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  • $\begingroup$ But this is assuming that the product in the category of sets is the same as the product in the category of sets with exactly 17 elements. Why can I make that assumption? $\endgroup$ – AmagicalFishy Jul 23 '15 at 0:31
  • $\begingroup$ No, I don't assume this. Recall that $X \times Y$, in any category, is defined by $\hom(-,X \times Y) \cong \hom(-,X) \times \hom(-,Y)$. $\endgroup$ – Martin Brandenburg Jul 23 '15 at 9:24
  • $\begingroup$ Oh! I didn't know this. Quickly looking up some notation, you're using hom functors, yes? (I haven't seen these yet—the exercise I was working on was in the first chapter of Mathematical Physics by Geroch; $X \times Y$ has been defined only in terms of the Cartesian Product between sets.) $\endgroup$ – AmagicalFishy Jul 23 '15 at 12:41
  • $\begingroup$ You don't have to know about hom-functors. If $(X \leftarrow P \rightarrow Y)$ is a product, you can directly verify that there is a bijection $\hom(T,P) \cong \hom(T,X) \times \hom(T,Y)$ for all objects $T$. (Actually this is natural in $T$, but this is irrelevant for the exercise.) $\endgroup$ – Martin Brandenburg Jul 23 '15 at 13:00
  • $\begingroup$ Oh. $hom(T,P)$ just refers to the "set" of morphisms from $T$ to $P$? What is $hom(T, X) \times hom(T, Y)$ defined categorically? (I know it's something like the Cartesian Product set-wise, but do all categories have an equivalent?) $\endgroup$ – AmagicalFishy Jul 23 '15 at 15:01

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