I know almost for all the examples such as ${\mathrm{Set}}$, $\mathrm{Cat}$, e.t.c., with the categorical product being the monoidal tensor product, the unit object is the terminal object. Is this an assumption, or it can be proved that if $(\mathcal{C},\times,I)$ is a monoidal category where $\mathcal{C}$ contains a terminal object, then $I$ has to be terminal? (I tried to prove this but failded.)
If this is only an assumption, is there any other "counterexample"?
Why do we need the unit object to be the terminal object in a cartesian monoidal category
category-theory
Related Solutions
No. For instance, taking the category of sets with the Cartesian product, the unit object (a singleton) is not integral since it has no morphism to the empty set. More generally, if you take the category of sheaves (of sets, or of abelian groups if you want an abelian category say) on some site, both properties (1) and (2) will typically fail since a sheaf can have no global sections or not be generated by global sections. For instance, if you take the category of sheaves of sets on $\mathbb{R}$ and let $X$ be the sheaf given by $X(U)=\{*\}$ if $U\subseteq(0,1)$ and $X(U)=\emptyset$ otherwise, the two inclusions $X\to X\coprod X$ are distinct but $X$ has no morphisms from the unit object (which is the constant singleton sheaf).
To define the product functor we need to fix a binary product, together with the projections, for each pair of objects.
The first thing to verify here is that it indeed defines a functor, i.e. uniquely determines the image of any pair of morphisms (in a functorial way).
I guess, the most problematic parts of the statement are the coherence conditions.
To get the proof smoother, let's also fix ternary and quarternary products (and nullary product as fixing the terminal object $1$, and unary products with the identity as projection).
For example, if we write $(a\times b\times c)\times d$, that is a binary product whose first term is a ternary product, and by the universal property we can easily see that it has a unique isomorphism from the quarternary product $a\times b\times c\times d$ that commutes with our fixed projections.
Similarly, for any act of 'putting in a pair of parenthesis' into a product of objects there belongs a unique isomorphism.
Now we define the associators $\alpha_{a,b,c}$ as the composite $(a\times b)\times c\longleftarrow a\times b\times c\longrightarrow a\times (b\times c)$ of the above isomorphisms.
Once again using the universal property of products, one can prove that if we put two pairs of parentheses in a product through the above isomorphisms, then the order doesn't matter: we'll receive commutative squares (like in the picture below).
To conclude the coherence pentagon condition, I found this picture among my old notes (on Leinster's unbiased bicategories), where the product sign $\times$ is omitted (so for example the top arrow is read as $((a\times b)\times c)\times d\overset{\alpha_{a,b,c}\times d}\longrightarrow (a\times(b\times c))\times d$).
The coherence conditions for unit are similar but easier, they might be thought as the case when we put an empty pair of parentheses into a product, like in "$ab\to ab()$" which would mean $a\times b\longrightarrow a\times b\times 1$.
Best Answer
Let $T$ be a terminal object for $\mathcal{C}$. Then $T\times I\cong T$ since $I$ is the unit of your monoidal structure, but also $T\times I\cong I$ since $T$ is terminal. So, $I\cong T$ and $I$ is also terminal.
(This is just the usual argument that the unit of a monoid is unique, applied to the monoid of objects of $\mathcal{C}$ up to isomorphism with $\times$ as the operation.)