# What is the relationship between external and internal composition in a cartesian closed category?

Denote "the" category of sets and functions by $$S$$. The hom set of functions from set $$X$$ to set $$Y$$ is denoted by $$S(X,Y)$$.

If $$C$$ is a cartesian closed category denote by $$C(x,y)$$ the set of morphisms from $$x$$ to $$y$$ in $$C$$. In such a $$C$$ there exists a natural bijection between $$C(x,y)$$ and $$C(1,y^x)$$. In a sense, $$y^x$$ reifies inside $$C$$ the set $$C(x,y)$$ in $$S$$. Both $$S(1,C(x,y))$$ and $$C(1,y^x)$$ are sets, and in particular, if $$x=1$$, then both $$S(1,C(1,y))$$ and $$C(1,y^1)$$ are sets.

Anyhow, how does the "external" law of composition $$C(x,y) \times C(y,z) \to C(x,z)$$ in $$S$$ of $$C$$ relate to the "internal" law of composition $$y^x \times z^y \to z^x$$ in $$C$$? In summary, does the internal composition "reify" the external composition?

• I've edited to add MathJax markup to your question. – Alex Kruckman Jul 6 at 17:53

The internal composition $$y^x\times z^y\to z^x$$ induces, by composition, $$C(1,y^x\times z^y)\to C(1,z^x)$$. The domain of this morphism is naturally equivalent to $$C(1,y^x)\times C(1,z^y)$$, by definition of product in $$C$$. So we get a function $$C(1,y^x)\times C(1,z^y)\to C(1,z^x)$$, which, as you noted, is equivalent to $$C(x,y)\times C(y,z)\to C(x,z)$$. And that's the external composition.