The binary product of two $\lambda$-presentable objects (in a locally presentable category) is $\mu$-presentable for some regular cardinal $\mu \geq \lambda$ (because all objects are $\mu$-presentable for some regular cardinal $\mu$). I don't see any reason for $\mu$ to be equal to $\lambda$. Even if $\lambda^2=\lambda$ and even if the binary product of two finite sets is finite for example.



> 
> Is there one ? Or is there a regular cardinal big enough $\mu$ such
>  that the class of $\mu$-presentable objects is closed under binary
>  product ?
> 
> 
> 



The binary product of two presentable objects
Unfortunately, no example for the first question is coming to mind at the moment.


For the second question, let $\lambda$ be such that the product functor $\times: \mathcal C \times \mathcal C \to \mathcal C$ is $\lambda$-accessible [1]. Let $\mu \rhd \lambda$ be such that the binary product of $\lambda$-presentable objects is $\mu$-presentable. I claim that $\mu$-presentable objects are closed under binary products.


To see this, let $A,B$ be $\mu$-presentable. Then $(A,B)$ is $\mu$-presentable in $\mathcal C \times \mathcal C$ [2]. Write $(A,B) = \varinjlim\_{i \in I} (A\_i, B\_i)$ where $I$ is $\lambda$-filtered and $\mu$-small [3]. Because $\times$ is $\lambda$-accessible, we have 


$$A \times B = \varinjlim\_I A\_i \times B\_i \qquad (\ast)$$


If $C = \varinjlim\_{j \in J} C\_j$ is a $\mu$-filtered colimit, then we have


$$\mathcal C(A \times B, C) = \mathcal C(\varinjlim\_{i \in I} A\_i \times B\_i, \varinjlim\_{j \in J} C\_j) = \varprojlim\_{i \in I} \mathcal C(A\_i \times B\_i, \varinjlim\_{j \in J} C\_j) \\
= \varprojlim\_{i \in I} \varinjlim\_{j \in J} \mathcal C(A\_i \times B\_i, C\_j) \\
= \varinjlim\_{j \in J} \varprojlim\_{i \in I} \mathcal C(A\_i \times B\_i, C\_j) \\
= \varinjlim\_{j \in J} \mathcal C(\varinjlim\_{i \in I} A\_i \times B\_i, C\_j) 
 = \varinjlim\_{j \in J} \mathcal C(A \times B, C\_j) $$


as desired. In the first line, we have used $(\ast)$. In the second line, we have used that $A\_i \times B\_i$ is $\mu$-presentable and $J$ is $\mu$-filtered. In the third line we have used that $I$ is $\mu$-small and $J$ is $\mu$-filtered. In the fourth line, we have used $(\ast)$ again.




---


[1] In fact any $\lambda$ will do here: recall that a right adjoint functor is $\lambda$-accessible iff its left adjoint preserves $\lambda$-presentable objects. The left adjoint of $\times$ is the diagonal functor $\Delta: \mathcal C \to \mathcal C \times \mathcal C$, which preserves $\lambda$-presentable objects for every $\lambda$.


[2] To see this, suppose that $A = \varinjlim\_{k \in K} A\_k$ and $B = \varinjlim\_{l \in L} B\_l$ where $K,L$ are $\mu$-small and $\lambda$-filtered. Then we claim that $(A,B) = \varinjlim\_{(k,l) \in K \times L} (A\_k, B\_l)$; then we can take $I = K \times L$, which is also $\mu$-small and $\lambda$-filtered. It suffices to show that the projection $\pi: K \times L \to K$ is cofinal (and dually). Now, for $k \in K$, $k \downarrow \pi = (k \downarrow K ) \times L$, which is a product of filtered categories and so filtered itself, and in particular nonempty and connected. So indeed $\pi$ is cofinal.


[3] Thanks to Mike Shulman, who points out below that this is Rmk 2.15 in Adámek and Rosický -- and rather, $A \times B$ is a *retract* of such a colimit -- but this is sufficient for our purposes.


