I am learning some basics from Category Theory and I am currently working on understanding Tensor-hom adjunction. I was reading a Wiki article about it (https://en.wikipedia.org/wiki/Tensor-hom_adjunction), but I am confused about the following:
When we construct a unit and counit as there as following: The unit
$$\varepsilon : FG \to 1_{\mathcal C}$$
has components
$$\varepsilon_Z : Hom_S (X,Z) \otimes_R X \to Z$$
given by evaluation:
\begin{align*}
& \phi \in Hom_S(X,Z) \quad \textrm{i} \quad x\in X \\
& \varepsilon(\phi \otimes x) = \phi(x).
\end{align*}
And counit with components
\begin{align*}
& \eta: 1_{\mathcal D} \to GF \\
& \eta_Y : Y \to Hom_S(X, Y \otimes_R X)
\end{align*}
we define as
$$\eta_Y(y)(t) = y \otimes t \quad \textrm{for} \quad t \in X.$$
Counit and unit equations can now be easily checked. What about naturality? How to formally prove that these transformations are indeed natural? Also, I was wondering if it is true that evaluation is always a natural transformation.
My work on proving $\varepsilon$ is natural:
To prove this, we have to show the following diagram commutes:
But, I am not sure what exactly $FG(f)$ is (it's defined for objects, but what about arrows?) and how to prove that this commutes.
Edit: So, we have functors defined as
\begin{align*}
F(Y) &= Y \otimes_R X \quad \textrm{za} \quad Y \in \mathcal D \\
G(Z) &= Hom_S(X,Z) \quad \textrm{za} \quad Z \in \mathcal C
\end{align*}
And what we are looking for is $FG(f)$ where $f: Z \to Y$. In the definition of functor, we have that if $f: X \to Y$ then $F(f): F(X) \to F(Y)$. So it is not hard to conclude that
$$G(f) : Hom_S (X,Z) \to Hom_S(X,Y)$$
and
$$FG(f) : Hom_S(X,Z) \otimes_R X \to Hom_S(X,Y) \otimes_R X$$
Is this enough to prove that such a diagram commutes?
Best Answer
To put our discussion in the comments into an answer. $\newcommand{\Hom}{\operatorname{Hom}}$
First of all, let's write down how these functors act on morphisms. Let $f\colon A\to B$. Then, $\Hom(X,f)\colon \Hom(X,A)\to \Hom(X,B)$. So, we take some morphism $s\colon X\to A$ and need to produce morphism $\Hom(X,f)(s)\colon X\to B$ which we can do easily with composition $X\stackrel{s}{\to} A \stackrel{f}{\to} B$, i.e. $\Hom(X,f)(s) = f\circ s$.
For tensor product, note that given any two linear maps $f\colon A\to A'$ and $g\colon B\to B'$, map $h\colon A\times B\to A'\otimes B'$ defined by $h(a,b) = f(a)\otimes g(b)$ is bilinear, so by the universal property there is a unique linear map $f\otimes g\colon A\otimes B\to A'\otimes B'$ such that $(f\otimes g)(a\otimes b) = f(a)\otimes g(b)$. So, given $f\colon A\to B$, we need $f\otimes X\colon A\otimes X\to B\otimes X$ which we can define simply as $f\otimes\mathrm{id}_X$.
We are now ready to check naturality:
\begin{array}{cccc} \Hom(X,A)&\!\!\!\!\otimes&\!\!\!\! X &\stackrel{\epsilon_A}{\longrightarrow} &A &\stackrel{f}{\longrightarrow} &B\\ \qquad\qquad s&\!\!\!\!\otimes&\!\!\!\! x &{\mapsto} &s(x) &{\mapsto} & f(s(x)) \end{array}
\begin{array}{cccc} \Hom(X,A)&\!\!\!\!\otimes&\!\!\!\! X &\stackrel{(f\circ-)\otimes\mathrm{id}_X}{\longrightarrow} &\Hom(X,B)&\!\!\!\!\otimes&\!\!\!\! X &\stackrel{\epsilon_B}{\longrightarrow} &B\\ \qquad\qquad s&\!\!\!\!\otimes&\!\!\!\! x &{\mapsto} &\qquad(f\circ s)&\!\!\!\!\otimes&\!\!\!\! x &{\mapsto} & (f\circ s)(x) \end{array}
which are obviously equal.