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?

3 Answers 3


$\require{AMScd}$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 $b$).

  • $\begingroup$ 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. $\endgroup$
    – PatrickR
    Sep 15, 2022 at 3:41
  • $\begingroup$ Yes actually: hadn't thought of that; thanks! In fact, it's enough for the monoid to be non-trivial. $\endgroup$ 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.

  • $\begingroup$ For the record, this is Exercise 1.2.25 in Leinster. $\endgroup$
    – PatrickR
    Sep 16, 2022 at 7:01

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