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Todd Trimble
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The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$. (And the hom-set of enriched transformations would be $\hom_C(1, 1)$, which is a 1-element set.) More generally still, if $C$ is symmetric monoidal closed and $I$ is the monoidal unit, the set of enriched transformations from $1_C$ to itself will be in natural bijection with $\hom_C(I, I)$; this specializes to Chris's observation where we have $\hom_{Vect}(k, k) \cong k$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.

Edit: Of course, I still haven't given you an example of a cartesian closed category where the identity functor has more than one self-transformation (in the unenriched sense). But these are easy to come by. Consider for example the topos $Set^G$ of permutation representations of an abelian group $G$, i.e., functors $G \to Set$ where $G$ is regarded as a one-object category. Then for any $g \in G$, the action $g \cdot - : X \to X$ on $G$-sets $X$ is a $G$-set morphism that provides a natural transformation from the identity $Set^G \to Set^G$ to itself.

The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$. (And the hom-set of enriched transformations would be $\hom_C(1, 1)$, which is a 1-element set.) More generally still, if $C$ is symmetric monoidal closed and $I$ is the monoidal unit, the set of enriched transformations from $1_C$ to itself will be in natural bijection with $\hom_C(I, I)$; this specializes to Chris's observation where we have $\hom_{Vect}(k, k) \cong k$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.

The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$. (And the hom-set of enriched transformations would be $\hom_C(1, 1)$, which is a 1-element set.) More generally still, if $C$ is symmetric monoidal closed and $I$ is the monoidal unit, the set of enriched transformations from $1_C$ to itself will be in natural bijection with $\hom_C(I, I)$; this specializes to Chris's observation where we have $\hom_{Vect}(k, k) \cong k$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.

Edit: Of course, I still haven't given you an example of a cartesian closed category where the identity functor has more than one self-transformation (in the unenriched sense). But these are easy to come by. Consider for example the topos $Set^G$ of permutation representations of an abelian group $G$, i.e., functors $G \to Set$ where $G$ is regarded as a one-object category. Then for any $g \in G$, the action $g \cdot - : X \to X$ on $G$-sets $X$ is a $G$-set morphism that provides a natural transformation from the identity $Set^G \to Set^G$ to itself.

added observations on the enriched case
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Todd Trimble
  • 53.3k
  • 6
  • 205
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The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$. (And the hom-set of enriched transformations would be $\hom_C(1, 1)$, which is a 1-element set.) More generally still, if $C$ is symmetric monoidal closed and $I$ is the monoidal unit, the set of enriched transformations from $1_C$ to itself will be in natural bijection with $\hom_C(I, I)$; this specializes to Chris's observation where we have $\hom_{Vect}(k, k) \cong k$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.

The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.

The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$. (And the hom-set of enriched transformations would be $\hom_C(1, 1)$, which is a 1-element set.) More generally still, if $C$ is symmetric monoidal closed and $I$ is the monoidal unit, the set of enriched transformations from $1_C$ to itself will be in natural bijection with $\hom_C(I, I)$; this specializes to Chris's observation where we have $\hom_{Vect}(k, k) \cong k$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.

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Todd Trimble
  • 53.3k
  • 6
  • 205
  • 322

The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.