Let $({\cal C}, \otimes, \cal I)$ be a monoidal (locally small) category and consider $\cal Hom : \cal C^{op} \times C \to Set$. The $\cal Hom$ functor is always a lax functor in the sense that $\forall A,B,C, D \in {\cal C}$ there is some natural transformation $\cal Hom(A,C) \times Hom(B,D) \to Hom(A \otimes B, C \otimes D)$. Then, what about categories in which this functor is strong monoidal (which means that the previous natural transformation is an isomorphism) ?

I conjecture that in this case the tensor should be a biproduct with $\cal I$ a zero object but I can't find any proof of this.

  • 1
    $\begingroup$ Biproducts don't work, since there you would expect to get factors $\hom(A, D)$ and $\hom(B, C)$ as well (think of the case $\mathcal{C} = \text{Vect}$). $\endgroup$
    – Todd Trimble
    Jul 22 '13 at 2:46
  • $\begingroup$ It is more natural to ask this for internal homs (if they exist) and $\otimes$ instead of $\times$. For example, if $A,B$ are dualizable objects of a closed symmetric monoidal category, then $\underline{\hom}(A,C) \otimes \underline{\hom}(B,D) \to \underline{\hom}(A \otimes B,C \otimes D)$ is an isomorphism. $\endgroup$ Aug 6 '13 at 7:20

It seems any such monoidal category $\mathcal{C}$ must be equivalent to the terminal category.

Let $I$ be the monoidal unit. First I claim there is exactly one morphism $I \to I$. For we have an isomorphism

$$\hom(A, B) \cong \hom(I \otimes A, I \otimes B) \cong \hom(I, I) \times \hom(A, B)$$

where the first is induced from unit isomorphisms $I \otimes A \cong A$, $I \otimes B \cong B$, and the second is inverse to the isomorphism assumed in the question. This isomorphism must take $f: A \to B$ to $(1_I, f)$, because one may check that its inverse takes $(1_I, f)$ to $f$. In particular, taking $A = B$ where $\hom(A, A)$ is nonempty, the composite

$$\hom(A, A) \stackrel{\cong}{\to} \hom(I, I) \times \hom(A, A) \stackrel{\text{proj}}{\to} \hom(I, I)$$

is surjective, but all $A \to A$ get mapped to $1_I$, so this is the only element of $\hom(I, I)$.

Second, I claim that any map $f: A \to B$ can be expressed as a composite $A \stackrel{f_1}{\to} I \stackrel{f_2}{\to} B$. If that is the case, it would apply to $f = 1_A$ where $f_2 \circ f_1 = 1_A$, but we also have $f_1 \circ f_2 = 1_I$ from what we just proved. Hence every object $A$ is isomorphic to $I$, whence $\hom(A, B) \cong \hom(I, I)$ is a singleton for every pair $(A, B)$, proving the statement in the first paragraph.

Proof of second claim: we have an isomorphism

$$\hom(A, B) \cong \hom(A \otimes I, I \otimes B) \cong \hom(A, I) \times \hom(I, B)$$

sending $f: A \to B$ to $(f_1: A \to I, f_2: I \to B)$ say, where $f: A \to B$ is retrieved as an evident composite

$$A \cong A \otimes I \stackrel{f_1 \otimes f_2}{\to} I \otimes B \cong B$$

where the displayed isomorphisms are unit isomorphisms. WLOG, we may simplify calculations by assuming that $\mathcal{C}$ is strict monoidal, where the unit isos are identities, and we have the crucial Eckmann-Hilton interchange

$$f = f_1 \otimes f_2 = (1_I \circ f_1) \otimes (f_2 \circ 1_I) = (1_I \otimes f_2) \circ (f_1 \otimes 1_I) = f_2 \circ f_1,$$

which proves the claim.

  • $\begingroup$ Thank you very much for your explanation, seems I was too optimistic. In fact, I was searching some necessary condition for $hom$ to have $hom(A \otimes B, I) \cong hom(A,I) \times hom(B, I)$ as well as $hom(I, A \otimes B) \cong hom(I, A) \times hom(I, B)$ and asking for $hom$ to be strong monoidal is a little too much as shown by your explanation. $\endgroup$ Jul 22 '13 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.