# What's the difference between a cartesian monoidal category and a semicartesian monoidal category?

In a semicartesian monoidal category, any tensor product of objects $x \otimes y$ comes equipped with morphisms $$p_x : x \otimes y \to x$$ $$p_y : x \otimes y \to y$$ given by $$x \otimes y \stackrel{1 \otimes e_y}{\longrightarrow} x \otimes I \stackrel{r_x}{\longrightarrow} x$$ and $$x \otimes y \stackrel{e_x \otimes 1}{\longrightarrow} I \otimes y \stackrel{\ell_y}{\longrightarrow} y$$ respectively, where $e$ stands for the unique morphism to the terminal object and $r$, $\ell$ are the right and left unitors. We can thus ask whether $p_x$ and $p_y$ make $x \otimes y$ into the product of $x$ and $y$. If so, it is a theorem that $C$ is a cartesian monoidal category. (This theorem is probably in Eilenberg and Kelly's paper on closed categories, but they may not have been the first to note it.) This also follows if we posit the existence of a natural diagonal morphism $x \to x \otimes x$.

Do $p_x$ and $p_y$ not make it into a full-fleged product? What am I missing? Is it that the extra $1 \otimes e_x$/$1 \otimes e_y$ arrows mean that $x \rightarrow p_x (x \otimes x)$ is merely an isomorphism and not the identity?

• The problem is probably that the projections $p_x$ and $p_y$ do not make $x \otimes y$ into the product, ie. the induced map $x \otimes y \xrightarrow{(p_x, p_y)} x \times y$ is not an isomorphism. I haven't worked out the examples they claim aren't cartesian, but it's what I would guess. – Najib Idrissi Sep 28 '14 at 11:21
• The nlab explicitly mentions many examples of semicartesian monoidal categories which are not cartesian. And I agree with Najib: The product of $x,y$ is more than just two morphisms $(p \to x, p \to y)$, the universal property has to be satisfied. – Martin Brandenburg Sep 28 '14 at 11:25
• Thanks; I was rather confused. – Sophie Taylor Sep 28 '14 at 11:43
• By $x → p_x (x⊗x)$, do you mean a lambda function which constructs the pair $(x⊗x)$, and then applies $p_x$ to it in order to get back $x$? You cannot write that in a semicartesian category, since there is no diagonal morphism $x → x⊗x$. – gelisam Nov 6 '18 at 1:00