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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.

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    $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Aug 6 '13 at 7:20
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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.

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  • $\begingroup$ 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. $\endgroup$ – kyo dralliam Jul 22 '13 at 9:59

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