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I'm trying to work my way through a category theory textbook on my own (Awodey's) and came across a problem asking about groups in a slice category $\mathbf{Sets}/I$ for any set $I$. For a group to exist in a category, Awodey said it must have finite products, which makes sense, since we need products to define a binary operation. Which led me to think about what a product is in $\mathbf{Sets}/I$

I figured the obvious candidate would likely be an arrow whose domain is the product in $\mathbf{Sets}$ but that doesn't seem to work, as follows.

Let $I$ be the set $\{1,2\}$, then consider the sets $\{1\}$, $\{2\}$. For objects in the slice category, take functions $i,j$ which will be the embeddings $\{1\}\rightarrow \{1,2\}$ and $\{2\}\rightarrow \{1,2\}$ respectively. Suppose there is a product $i\times j$ which has as its domain the set $\{(1,2)\}$. Let $\bar 1,\bar 2:\{(1,2)\}\rightarrow \{1,2\}$ denote the constant functions which map to 1 and 2 respectively.

Then, $i\circ \bar 1:\{(1,2)\}\rightarrow \{1,2\}$ is the constant function which maps to 1, and $i\circ\bar 2$ is the constant function which maps to 2. There's no single choice for $i\times j$ which makes it an actual candidate to be a product. (I also don't think disjoint union works, it doesn't satisfy the UMP I don't think.)

So then, what is a product in $\mathbf{Sets}/I$, and if it is some function from the product in $\mathbf{Sets}$ where's the mistake in my reasoning above?

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The product in a slice category is the fiber product or pullback. Thinking of the slice category over a set $I$ as the category of $I$-indexed sets, it is the "pointwise" product. In your example, the product is empty.

More explicitly, if $f : X \to I$ and $g : Y \to I$ are objects in the slice category of sets over $I$, then their product is

$$X \times_I Y = \{ (x, y) : f(x) = g(y) \}$$

where the function to $I$ is the common value of $f$ and $g$. More evocatively, writing $X_i = f^{-1}(i), i \in I$ and similarly for $Y$, we have

$$(X \times_I Y)_i = X_i \times Y_i.$$

As a side comment, it is in fact not necessary for a category to have products to define group objects in it. Here is a definition which does not require the existence of products and which reduces to the usual definition if products exist:

A group object in a category $C$ is a functor $C^{op} \to \text{Grp}$ whose underlying functor to $\text{Set}$ is representable.

Basically the point is that you can instead take products in presheaves over $C$, which always exist, and which agree with products in $C$ under the Yoneda embedding if those exist. For example, by this definition the cyclic group $C_2$ of order $2$ is meaningfully a group object in the category of sets of size at most $2$, even though the product $C_2 \times C_2$ doesn't exist.

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    Is there a way in general to come up these, or is this, like finding counterexamples, more of something to play with and try until finding something that works? Also, thanks, that's very helpful! Though I'm afraid that as I just started learning category theory, the entire paragraph about that alternative definition of groups in a category was rather incomprehensible. Underlying functor to Set? Presheaves, Yoneda embedding? Guess I have a lot to learn.2017-01-19
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    @drowdemon: you can try paying very careful attention to exactly what the universal property you're trying to satisfy is. If you write it down you find that the product of $f : X \to I$ and $g : Y \to I$ in $\text{Set}/I$ is a set $P$ and a pair of maps $a : P \to X, b : P \to Y$ such that $f \circ a = g \circ b$ which is universal with respect to this property. This is precisely the definition of a pullback, although if you didn't know that you could still figure out what set $P$ has to be by staring carefully at this; the condition that $f \circ a = g \circ b$ is key.2017-01-19
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    The last bit was mostly a side comment; it wasn't intended to be comprehensible to someone just learning category theory. Sorry about that!2017-01-19