Difference between internal hom and exponential objects?

I'm reading that exponential objects are only "examples" of internal homs. (I used to think they were just synonyms).

I'm reading the definition of internal hom, and I cannot really understand the point of it.

What is the difference between internal homs and exponential objects? i.e. what does the internal hom mean, in cases where it doesn't mean exponential object?

• according to the n category lab, an exponential object is an internal hom in a category that's cartesian closed. ncatlab.org/nlab/show/exponential+object Apr 5 '19 at 15:07

I don't know exactly which sources and definitions you are looking at, but usually the distinction is the following:

An Internal Hom $$[X,-]$$ is the right adjoint to the functor $$- \otimes X$$ given by a tensor product.

An Exponential Object is the specific case of an internal hom where the tensor product is given by the cartesian product $$- \times X$$.

There are a good number of interesting tensor products that aren't the cartesian product, so we get many examples of internal homs which aren't exponential objects.

One good example comes from the category of vector spaces. We want the internal hom $$[V,W]$$ to be the space of linear maps from $$V$$ to $$W$$. To see what this should be the right adjoint of, let's consider linear maps $$U \rightarrow [V,W]$$.

If this internal hom was going to be an exponential object, we'd hope that these would correspond to linear maps $$U \times V \rightarrow W$$. However, this is not the case. Instead, each linear map $$U \rightarrow [V,W]$$ corresponds to a bilinear map $$U \times V \rightarrow W$$. So it's not an exponential object according to the definition.

How is $$[V,W]$$ an internal hom according to the definition? The answer to this gives us our first interesting example of a tensor product that isn't cartesian: the tensor product of vector spaces $$U \otimes V$$. There's a number of ways of defining it (for finite dimensional spaces, it has dimension $$dim(U \otimes V) = dim(U) \times dim (V)$$) but the key point is that linear maps $$U \otimes V \rightarrow W$$ correspond to bilinear maps $$U \times V \rightarrow W$$. Which, as we've already seen, correspond to linear maps $$U \rightarrow [V,W]$$.

In other words, the space of linear maps $$V \rightarrow W$$ is an internal hom, but the tensor product it corresponds to is the vector space tensor product, not the cartesian one.

• VERY clarifying thanks! Apr 5 '19 at 15:26
• I guess I just have a mental block about this, but I'm really confused. Why are we considering linear maps $U\to [V,W]$? And why and in what sense do we expect them to correspond to linear maps $U\times V \to W$? I'm sorry for the dumb question, but if you're able to clarify it would be appreciated! Apr 19 '20 at 16:11
• @Nathaniel It's a statement of what we would need for the functors $- \times V$ and $[V, -]$ to be left and right adjoints. (Of course they aren't adjoints because the linear maps don't actually correspond that way, which shows that the cartesian product is the wrong product to go with the internal hom. But if we weren't aware of the tensor product of vector spaces and thought the cartesian product was the only one available, that correspondence is what we'd check.) Apr 19 '20 at 17:52