# Functoriality of bifunctors: Joint functoriality equivalent to separate functoriality?

Since I haven't found anything related on MSE nor in Google, I'll post here this question.

Let $$\mathsf{C}, \mathsf{D}, \mathsf{E}$$ be categories and consider an assignment $$F:\mathsf{C}\times\mathsf{D}\to\mathsf{E}$$. Then I'm wondering whether $$F$$ is functorial iff it's functorial on $$\mathsf{C}$$ and on $$\mathsf{D}$$. I'll define what I mean by "functorial in a variable". For an object $$d\in\mathsf{D}$$, define the assignment \begin{align*} F_d:\mathsf{C}&\to\mathsf{E}\\ c&\mapsto F(c,d)\\ (f:c\to c')&\mapsto F(f,1_d). \end{align*}

And for each object $$c\in\mathsf{C}$$, define $$F^c:\mathsf{D}\to\mathsf{E}$$ similarly. Is it true then that $$F$$ is a functor iff $$F^c$$ and $$F_d$$ are functors for all objects $$(c,d)\in\mathsf{C}\times\mathsf{D}$$? In other words: is joint functoriality equivalent to separate functoriality? The implication to the right is easy, and for the implication to the left, it's not difficult to show that $$F$$ will preserve identities using that $$F_d$$ preserves identities. What I'm having trouble to show is that $$F$$ does preserve compositions. Maybe it's false? I cannot come with a counterexample right now.

Edit: It's false, separate functoriality alone does not imply joint functoriality. I've found the following counterexample: define $$\mathsf{C}$$ to be the free category over the quiver $$\bullet\to\bullet$$. That is, $$\mathsf{C}$$ has two objects, $$a$$ and $$b$$, and three morphisms, $$1_a$$, $$1_b$$ and $$f:a\to b$$.

Now define the assignment $$F:\mathsf{C}\times\mathsf{C}\to\mathsf{C}$$, for $$g,h\in\operatorname{Mor}\mathsf{C}$$, as $$F(g,h)= \begin{cases} 1_a,&g=h=f,\\ 1_b,&\text{otherwise}, \end{cases}$$ and, implicitly, $$F(c,d)=b$$ for $$c,d\in\{a,b\}$$. Then $$F_c$$ and $$F^c$$ are functors for $$c\in\{a,b\}$$ (namely, they are the constant functor $$\mathsf{C}\to\mathsf{C}$$ to $$b$$). So $$F$$ is separately functorial. But it is not jointly functorial, since although $$(1_b,1_b)$$ is composable with $$(f,f)$$ in $$\mathsf{C}\times\mathsf{C}$$, we have that $$F(1_b,1_b)=1_b$$ is not composable with $$F(f,f)=1_a$$ in $$\mathsf{C}$$.

Note that the assignment of this counterexample does not respect domains, $$F(\operatorname{dom}(f,f))=b\neq a=\operatorname{dom}(F(f,f))$$.

So now the questions are:

1. Can an ‘easy’ counterexample still be found if require $$F$$ to respect domains and codomains?
2. Can some additional condition be added to separate functoriality to achieve equivalence with joint functoriality?

$$\require{AMScd}$$I don't remember what this condition is called (it had the name of a category theorist, or even two) but a bifunctor $$F : {\cal A}\times{\cal B} \to {\cal C}$$ is such that for $$f:A\to A', g:B\to B'$$, the diagram $$\begin{CD} F(A,B) @>>> F(A,B') \\ @VVV @VVV \\ F(A',B) @>>> F(A',B') \end{CD}$$ commutes. This is the sense in which the morphism $$F(f,g) : F(A,B) \to F(A', B')$$ is defined: it's either path of this commutative square. So, a family of functors $$F_A : {\cal B} \to {\cal C}, A\in\cal A$$ and a family of functors $$F_B : {\cal A} \to {\cal C}, B\in\cal B$$ are induced by a common bifunctor $$\bar F : {\cal A}\times{\cal B} \to {\cal C}$$ (in the sense that $$F_A = \bar F(A,-), F_B=\bar F(-,B)$$) if and only if for each $$f:A\to A', g:B\to B'$$ one has $$F_{B'}(f)\circ F_A(g) = F_{A'}(g)\circ F_B(f).$$

Let $$(D,E)$$ be the category of functors from $$D$$ to $$E$$ with morphisms given by transformations that omit the naturality condition, i.e. morphisms being transformations, i.e. families of morphisms $$\beta_d\colon Gd\to Hd$$, one for each object $$d\in D$$. Then $$F(-,-)\colon C\times D\to E$$ is an association (respecting domains and codomains) with each $$F^c$$ a functor if and only if we have an association (respecting domains and codomains) $$F^-\colon C\to(D,E)$$ given by $$F^c(d)=F(c,d)$$. This association $$F^-$$ is itself a functor if and only if additionally each $$F_d$$ is a functor.

Moreover, $$F$$ is a bifunctor if and only if $$F^-\colon C\to(D,E)$$ factors through the inclusion of $$[D,E]\hookrightarrow (D,E)$$, i.e. if and only if for each morphism $$f\colon c\to c'$$ in $$C$$, the transformation $$F^f\colon F^{c'}\Rightarrow F^c$$ with components $$F^f_d=F(f,1_d)\colon F^{c'}(d)=F(c',d)\to F(c',d)$$ is natural, i.e. satisfies $$F(f,1_d)F(c,g)=F(c',g)F(f,1_d')$$ for each morphism $$f\colon c\to c'$$ in $$C$$ and each morphism $$g\colon d\to d'$$.

Finally, since naturality is the statement that certain squares commute, the simplest example of naturality failing is given by a non-commutative square. Explcititly, let $$C$$ and $$D$$ each be the category with two objects and one non-identity arrow between them. Then $$C\times D$$ is a category consisting of four objects and non-identity morphisms assembled in a commutative square. Then an association $$F\colon C\to D\to E$$ (preserving domains and codomains) has $$F^c$$ and $$F_d$$ be functors if and only if it sends identity morphisms to identity morphisms, in which case its image is a non-commutative square, with identity morphisms on its verticaes. The association is a bifunctor if and only if the image of the commutative square is commutative.

The smallest category $$E$$ containing a non-commutative square has one object and three morphisms, i.e. is a monoid with two non-identity morphisms $$a$$ and $$b$$ such that $$ab\neq ba$$. Indeed, such a monoid structure is generated by $$ax=a$$ and $$bx=b$$ for all $$x$$ (this is associative because it's simply taking the leftmost-non-identity term in any expression of $$a$$ and $$b$$).

• Would a smaller category than $E$ with a non-commutative square be a monoid with one non-identity morphism $a$ (and $aa$ either equal to $a$ or to $e$, it does not matter)? If $e$ is the identity morphism, we can then go around one half of the square by $ee=e$ and around the other half by $ae=a$ and the square will not commute. Sep 15, 2022 at 3:41
• Yes actually: hadn't thought of that; thanks! In fact, it's enough for the monoid to be non-trivial. Sep 15, 2022 at 17:10

Okay, I will spell out the full characterization, with the condition stated in fosco's answer, and give some additional details involved in the proof.

Proposition. Let $$F:\mathsf{C}\times\mathsf{D}\to\mathsf{E}$$ be an assignment from the product category $$\mathsf{C}\times\mathsf{D}$$ to $$\mathsf{E}$$. That is, $$F$$ is an assignment between classes of objects and between classes of morphisms. Then $$F$$ is a functor if and only if the following conditions holds:

1. $$F^c$$ and $$F_d$$ are functors for every object $$c\in\mathsf{C}$$ and $$d\in\mathsf{D}$$.
2. $$F(f,1_{d'})\cdot F(1_c,g)=F(1_{c'},g)\cdot F(f,1_d)$$ for every morphism $$f:c\to c'\in\mathsf{C}$$ and $$g:d\to d'\in\mathsf{D}$$.
3. For morphisms $$f\in\mathsf{C}$$ and $$g\in\mathsf{D}$$, the morphism $$F(f,g)$$ equals any of the members of the equality in 2.

The fact that $$F$$ is functor implies 1, 2 and 3 should be clear.

For the converse, note first that since $$F^c$$ and $$F_d$$ are functors, they preserve domains and codomains, so that the morphisms in the equality of 2 are actually composable. Next, observe that $$F$$ preserves domains and codomains since $$F^c$$ and $$F_d$$ preserve domains and codomains (they do since they are functors): \begin{align*} \operatorname{dom}(F(f,g)) &=\operatorname{dom}(F(1_{c'},g)\cdot F(f,1_d))\\ &=\operatorname{dom}(F(f,1_d))\\ &=\operatorname{dom}(F_d(f))\\ &=F_d(\operatorname{dom}f)\\ &=F_d(c)\\ &=F(c,d)\\ &=F(\operatorname{dom}(f,g)). \end{align*} And $$\operatorname{codom}(F(f,g))=F(\operatorname{codom}(f,g))$$ is done similarly.

The fact that $$F$$ preserves units follows from the fact that $$F_d$$ preserves units, \begin{align*} F(1_{(c,d)})&=F(1_c,1_d)\\ &=F_d(1_c)\\ &=1_{F_d(c)}\\ &=1_{F(c,d)}. \end{align*}

Lastly, the fact that $$F$$ preserves compositions can be deduced from conditions 2 and 3.

From the proposition one obtains a

Corollary. There is a one-to-one correspondence \begin{align*} \newcommand{\testleftlong}{\longleftarrow\!\shortmid} \begin{Bmatrix} \text{Functors}\\ F:\mathsf{C}\times\mathsf{D}\to\mathsf{E} \end{Bmatrix} &\longleftrightarrow \begin{Bmatrix} \text{Collections of functors }\\ \{F_d:\mathsf{C}\to\mathsf{E},F^c:\mathsf{D}\to\mathsf{E}\}_{(c,d)\in\operatorname{Ob}(\mathsf{C}\times\mathsf{D})}\\ \text{such that }F_d(c)=F^c(d),\;\forall (c,d)\in\mathsf{C}\times\mathsf{D},\\ \text{and }F_{d'}(f)\cdot F^c(g)=F^{c'}(g)\cdot F_d(f),\\ \forall (f:c\to c',g:d\to d')\in\mathsf{C}\times\mathsf{D}. \end{Bmatrix}\\ &\\ F&\longmapsto\{F_d=F(-,d),F^c=F(c,-)\}_{(c,d)\in\mathsf{C}\times\mathsf{D}}\\ \begin{pmatrix} F:\mathsf{C}\times\mathsf{D}\to\mathsf{E}\\ (c,d)\mapsto F_d(c)=F^c(d)\\ (f,g)\mapsto F_{d'}(f)\cdot F^c(g) \end{pmatrix} &\testleftlong\{F_d,F^c\}_{(c,d)\in\mathsf{C}\times\mathsf{D}}\\ \end{align*}

In the corollary, the map to the left is well-defined thanks to the proposition. It is easy to verify that the compositions "first to the right, then to the left" and the vice versa are the identities.

• For the record, this is Exercise 1.2.25 in Leinster. Sep 16, 2022 at 7:01