# Example of a category without product

A question was raised in our class about the non-existence of product in a category. The two examples that came up in the discussion was the category of smooth manifolds with boundary and the category of fields. But I cannot formally proof why categorical product does not exist in these categories.

I can see that category of finite fields cannot have infinite products due to cardinality reasons, but I would like to know why category of fields cannot even have finite products.

Is there any other simple example of a category without (finite) products? Any help regarding this appreciated.

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Without this being a general rule, if a category $C$ itself has finite products, and you take a full subcategory in which some of those products are not present, there is a good chance you will be able to prove that the subcategory doesn't have products. Since the ring-theoretic product of fields is not a field, it was a good bet that fields wouldn't have products. There are counterexamples to this, but it does suggest a number of candidates for categories without products. – David Feb 3 at 7:22

For fields the most basic problem is fields of different characteristic. If $K$ and $L$ are fields, then the product $K\times L$ (if it existed) would have to be a field that can map into both $K$ and $L$, which means in particular it has the same characteristic as both $K$ and $L$, which is impossible if $K$ and $L$ have different characteristic.

Even if you restrict to fields of a fixed characteristic, however, you still don't have products. For instance, in the category of fields of some fixed characteristic, let $K$ be any field with a non-identity endomorphism $e:K\to K$. Then I claim there is no product $K\times K$. For suppose there were; call it $P$, and let $p:P\to K$ and $q:P\to K$ be the two projections. Considering $L=K$ and the identity maps $f=g=1:L\to K$, by the universal property of the product there must exist $h:L\to P$ such that $ph=f$ and $qh=g$. Since $f$ and $g$ are the identity and every map of fields is injective, this means that $h$ must be an isomorphism and $p=q$ is its inverse. But now consider $g'=e:L\to K$; there is then an $h':L\to K$ such that $ph'=f$ and $qh'=g'$. Since $p=q$, this means $f=g'$, which is a contradiction since we assumed $e\neq 1$.

The case of smooth manifolds without boundary is fairly subtle (indeed, the category of smooth manifolds has some limits you might not expect it to have). Let me sketch a proof that there is no product $[0,1)\times[0,1)$ in the category of smooth manifolds with boundary. Suppose you had such a product $P$. By considering maps from the $1$-point manifold to $P$, you can identify the underlying set of $P$ with the cartesian product $[0,1)\times[0,1)$. By considering smooth maps from $\mathbb{R}$, you can show that $P$ is connected. By considering smooth maps from $\mathbb{R}^2$, you can show that $P$ is $2$-dimensional and that $\{0\}\times [0,1)\cup[0,1)\times\{0\}$ must be its boundary (here we use that a point is in the interior of a $2$-manifold iff there is an injective smooth map from $\mathbb{R}^2$ to the manifold whose image contains the point but no such map from $\mathbb{R}^n$ for any $n>2$). Again using smooth maps from $\mathbb{R}$, you can show that this boundary must be connected, and that (via the classification of $1$-manifolds) in fact it must be diffeomorphic to $\mathbb{R}$. Such a diffeomorphism $\mathbb{R}\to P$ onto the boundary then gives a smooth injection $i:\mathbb{R}\to\mathbb{R}^2$ whose image is $\{0\}\times [0,1)\cup[0,1)\times\{0\}$, and $i$ has the property that every smooth map $j:\mathbb{R}\to\mathbb{R}^2$ whose image is contained in $\{0\}\times [0,1)\cup[0,1)\times\{0\}$ factors smoothly through $i$. You can show that no such $i$ exists by examining what it does near the corner: assuming without loss of generality that $i(0)=(0,0)$, then all the derivatives of $i$ must vanish at $0$. But then $j(x)=i(\sqrt[3]{x})$ is smooth and does not factor through $i$, which is a contradiction.

For other examples of categories without products, it is very easy to get examples if you look at small categories you think of as an individual algebraic structure, rather than categories like "the category of all structures of a certain kind". For instance, you can form a category with any set of objects and no non-identity morphisms, and this doesn't have products of distinct objects. Or you can take a monoid and consider it as a category with one object, and it will rarely have a product of two copies of the object.

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Could you please explain why there can be no injective smooth map from $\mathbb{R}^n\to P$ for $n > 2$? Also, I did not get your argument about the map $j$ not factoring through $i$ – ChesterX Feb 3 at 17:25
An injective smooth map $\mathbb{R}^n\to P$ is the same as two smooth maps $f,g:\mathbb{R}^n\to [0,1)$ such that $f(x)=f(y)$ and $g(x)=g(y)$ implies $x=y$. You can combine these into a single smooth map $(f,g):\mathbb{R}^n\to\mathbb{R}^2$ which is injective. No such map exists for $n>2$. The map $j$ does not factor through $i$ because if it did, the factoring map $\mathbb{R}\to\mathbb{R}$ would have to be $x\mapsto \sqrt[3]{x}$, which is not smooth. – Eric Wofsey Feb 3 at 19:11
And how are you getting that all derivatives of $i$ must vanish at $0$? I can see this will imply smoothness of $j$ and so $i^{-1} \circ j$ must be smooth : a contradiction. – ChesterX Feb 4 at 7:02
As you approach $(0,0)$ along $[0,1)\times\{0\}$, all the derivatives must be pointing horizontally, and you approach $(0,0)$ along $\{0\}\times[0,1)$, they must all be pointing vertically. So by continuity, at $(0,0)$, they must point both horizontally and vertically, which means they must vanish. – Eric Wofsey Feb 4 at 7:10

Let $K$ be a field. In the category of fields, let's prove $P = K \times K$ (categorical product) doesn't exist in general. If it did, it would come equipped with two projection morphisms $\pi_1, \pi_2 \colon P \to K$. If $\operatorname{id} \colon K \to K$ is the identity morphism, the pair $(\operatorname{id},\operatorname{id})$ must factor through $P$, showing that $\pi_1, \pi_2$ are surjective, hence are isomorphisms. Thus we may assume $P = K$ and $\pi_1 = \pi_2 = \operatorname{id}$.

Now if $L$ is an arbitrary field and $\phi_1,\phi_2 \colon L \to K$ are morphisms, then applying the universal property of $P$, there must be some $\psi \colon L \to P$ such that $\phi_1 = \pi_1 \circ \psi = \psi = \pi_2 \circ \psi = \phi_2$. Thus any two embeddings $L \to K$ must coincide.

In particular, when $L = K$, this implies that no field has any nontrivial automorphisms. This is absurd, as the example of complex conjugation on $\mathbb{C}$ shows.

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Posets (viewed as categories) provide many examples of categories without certain limits. A product of two elements in a poset is simply a greatest lower bound, so you just need to make a poset where some pair of elements has no greatest lower bound. For example, the poset $\{ a, b, c, d \}$ where $a \leq c, a \leq d, b \leq c, b \leq d$. Then both $a$ and $b$ are lower bounds for $c$ and $d$, but they are incomparable.

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Minimal example: in the discrete category on the two-object set $\{B,A\}$, the product $B \times A$ doesn't exist. This is most easily seen by thinking of $\{B,A\}$ as a discrete poset. – goblin Feb 3 at 12:21