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

Question

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?

Answer 1

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.