# Why do cartesian closed categories by definition have a terminal object?

I find it intuitive to define a class of categories that have categorical products and exponentials. However, it is not obvious to me why the additional requirement that they have a terminal object makes the class of categories significantly more interesting.

Why don't people instead place more importance in the class of categories that have products and exponentials but not necessarily terminal objects? What does the terminal object bring us?

Context: I'm very new to category theory.

• well, you want to have arbitrary finite products, so either you say you have binary products and something that essentially is an empty product (the terminal object), or you say any finite set has a product (which produces the terminal object as the product of the empty diagram). also, as soon as you want your product to be monoidal, you will get immediately that the unit of this monoidal structure is a terminal object. In the end it is similar to wanting a unit in a group in order to have better control on it. – Enkidu Jan 29 at 7:19
• @Enkidu Why are you writing an answer in the comments? – Arthur Jan 29 at 7:21
• It lacks too many details for me to accept is an answer myself – Enkidu Jan 29 at 7:22

## 1 Answer

In a Cartesian closed category, we want a morphism $$f:A\to B$$ to have some kind of transpose, a morphism with codomain $$B^A$$. But you only get such a morphism if there is an $$X$$ with $$A\times X\simeq A$$, and with a terminal object you get an $$X$$ that does this with every $$A$$. With that in place, every $$f:A\to B$$ corresponds to a unique morphism $$\hat{f}:1\to B^A$$.

Also intuitively, it seems like if you have a function $$f:A\to B$$ this should also induce a morphism from $$A$$ to the set of all constant functions in $$B^C$$ for any $$C$$. It would be nice to define constant functions by something like factoring though an object that's like the set of all functions to $$B$$ from a one-element domain. And we get something like this latter object if we have a terminal object and the exponential $$B^1$$.