What would a category product with a single object instead of two objects be?

Question

In my attempt to get an intuition over the definition of the category product:

particulary over the uniqueness requirement of f (now I kind of see that if there can be many f then the product object is "too big" and if there's no f the product would be "too small"), and since according to Wikipedia there can be products with more than two objects, I tried to make it simpler to myself by considering a product with a single object:

So in this case, the definition would be that for every f₁ there is an f such that g∘ f = f₁. It seems to me that if this is the case then g would be a kind of isomorphism, and I think that if g actually is an isomorphism then there will always be a single f meeting the requirements (by doing g⁻¹∘ f₁), but I'm not sure this implication also goes the other way.

So my question is, what would this tell us about g? Is this a thing in category theory?

Answer 1

First think about what the product of one thing means typically: if I were to write $\prod_{i=1}^1 x_i$, I'm multiplying all $1$ of the $x_i$s together, and I end up with just $x_1$. So, before thinking too hard we might guess that $X_1 \times = X_1$, or at least isomorphic.

Secondly, recall that the definition of a product actually involves three things: $X_1 \times X_2$, and the two projections $g_1: X_1 \times X_2 \to X_1$ and $g_2: X_1 \times X_2 \to X_2$. So, the definition of a 1-product should involve two things: $X_1 \times$ and $g: X_1 \times \to X$.

Finally, we hope that $g$ is an isomorphism between the two objects, so let's try to show this. First, use $Y=X$ and $f_1$ as the identity map to construct $h$ which satisfies $\mathrm{id}_X = g \circ h$.

Now, use $Y = X\times$ and $f_1$ as the map $g: X \times \to X$ giving there is a unique $p: X \times \to X \times$ such that $g = g \circ p$. But note that $g = g \circ \mathrm{id}_{X\times}$ and $g \circ (h \circ g) = g \circ h \circ g = \mathrm{id}_X \circ g = g$, so we must have that $h \circ g = \mathrm{id}_{X\times}$ by uniqueness.

Thus, $g$ is an isomorphism.

Answer 2

You are correct that an isomorphism gives the product of a single object, for the reason you described. The implication goes the other way (in the sense that if a map $g$ satisfies this universal property, it must be an isomorphism). This is a consequence of a very general fact about universal properties: if you have two different objects with morphisms satisfying a universal property, there is a unique isomorphism between them that respects these morphisms.

To see how this works out in your example, you know that there is a product of $X_1$ given by $X_1$ itself, with $g:X_1\to X_1$ the identity map. Now suppose that $g':Z\to X_1$ is another product of $X_1$. Because $X_1$ and $g$ satisfy the universal property, there is a unique map $f:Z\to X_1$ with $g\circ f = g'$. On the flip side, since $Z$ and $g'$ satisfy the universal property, there is a unique map $f':X_1\to Z$ with $g'\circ f' = g$.

Here we used the "existence" part of the universal property to get two maps $f$ and $f'$. Then we claim $f$ and $f'$ are inverses, which will use the "uniqueness" part. The trick is to use the universal property satisfied by the map $g:X_1\to X_1$ on $g$ itself! This implies that there is a unique map $h:X_1\to X_1$ such that $g\circ h = g$. This could just be the identity map, but by the above, it could also be $f\circ f'$, since $g\circ f\circ f' = g'\circ f' = g$. So by uniqueness, $f\circ f'$ is the identity. The same type of reasoning shows $f'\circ f$ is.

So all of the different products are related by unique isomorphisms. In particular, in the case you cite, $g$ is necessarily an isomorphism. This is a general feature of universal properties, which is why we talk about "the" product of some objects -- there may be many products, but they will all be isomorphic in unique ways.