# Naturality of $\varphi : \textbf{Vct}_K(V(x), w) \xrightarrow{\sim} \textbf{Set}(x, U(w))$ in the variable $x$ (Cats for the Working Mathematician).

Consider the forgetful functor $$U : \textbf{Vct}_K \to \textbf{Set}$$ and the functor $$V : \textbf{Set} \to \textbf{Vct}_K$$ that takes an object $$x$$ in set to the $$K$$-vector space $$V(x)$$ with basis $$x$$ (a type of formal generation) and takes a map $$h : x \to x'$$ to the $$K$$-linear map $$V(h)(\sum r_i x_i) \ (r_i \in K, x_i \in X) = \sum r_i h(x_i)$$.

If $$g: x \to U(w)$$ is a map of sets, then:

Each function $$g: x \to U(w)$$ extends to a unique linear transformation $$f: V(x) \to w$$, given explicitly by $$f(\sum r_i x_i) = \sum r_i g(x_i)$$. The inverse of $$\psi : g \mapsto (f = V(g))$$ is $$\varphi : f \mapsto f \vert_x$$ the restriction of $$f$$ to the basis set $$x$$. Thus we have components:

$$\varphi_{x, w} : \textbf{Vct}_K(V(x), w) \xrightarrow{\sim} \textbf{Set}(x, U(w))$$

(here $$\varphi_{x,w} = \varphi$$ from preceeding remark) of a family of bijections between such homsets. These bijections happen to be natural in $$x$$ and $$w$$ so we have an isomorphism of bifunctors.

I'm trying to prove naturality of $$\varphi$$ in the first argument $$x$$ by showing that the following diagram is commutative for any general $$h : x' \to x$$ in $$\textbf{Set}$$ (yes, $$x' \to x$$ is the correct direction because $$x$$ is covariantly present in the first (thus, domain-reversing) argument of $$\textbf{Vct}_K(V(\cdot), \cdot)$$):

$$\require{AMScd} \begin{CD} \textbf{Vct}_K(V(x), w) @>{\varphi_{x,w}}>> \textbf{Set}(x, U(w))\\ @V{(Vh)^*}VV @VV{h^*}V \\ \textbf{Vct}_K(V(x'), w) @>{\varphi_{x',w}}>> \textbf{Set}(x', U(w)) \end{CD}$$

where $$h^*(g) \equiv g \circ h$$, $$(Vh)^*(g') = g' \circ (Vh)$$, and I'm doing this by directly substituting in and expanding the above formulas.

We want to show that:

$$h^* \circ \varphi_{x,w} = \varphi_{x', w} \circ (Vh)^*$$

or equivalently that for any $$g:V(x) \to w$$ element in the upper left corner of the diagram we have:

$$[h^* \circ \varphi_{x,w}(g)](y) = [\varphi_{x', w} \circ (Vh)^*(g)](y)$$

for every $$y \in V(x)$$. Using the definition of $$h^*$$ etc we have that the above is equivalent showing:

$$[\varphi_{x,y}(g) \circ h](y) = [\varphi_{x', w} \circ g \circ (Vh)](y)$$

for every $$y \in V(x)$$. This is where confusion sets in for me.

You're very close; to show this you don't need to refer to explicit elements $$y \in V(x')$$. To finish this, you need to take take advantage of the universal properties, and interpret what $$\varphi_{x,w}$$ really does to the morphisms.
To fill in those details you can create a commutative diagram like this: This diagram might help. Here the $$i$$'s are inclusion morphisms. The last equation you wrote corresponds to the commutativity of the colored-in diagram; see if you can reproduce this diagram by writing explicitly what $$\varphi_{x,w}$$ does.
• Sorry I should clarify that; $i: x \to U(V(x))$ is "universal" in the sense that for any $f: x \to U(W)$ with $W$ a vector space, we have that a unique $g: V(x) \to W$ with the diagram commuting. Here, $\phi$ is a function the other way around: we take a $g: V(x) \to W$ to its unique $\phi_{x, w}(g): x \to U(W)$ such that $U(g) \circ i = \phi_{x, w}(g)$. You might already understand all this so sorry if that's overkill. Anyways, that's what I meant, so apologies if that was confusing. – trujello Aug 4 '20 at 20:46
• So to figure this out, I knew that we needed to figure out what exactly $\phi$ "does." But since I knew $i: x \to U(V(x))$ is universal, I deduced that $\phi$ must send each $g: V(x) \to W$ to the unique $f: x \to U(W)$ such that $U(g) \circ i = f$ (which exist by universality). Thus $\phi$ is defined to go back and forth between $g: V(x) \to W$ and $f: x \to U(W)$ s.t. $U(g) \circ i = f$, which are in unique correspondence by universality. – trujello Aug 4 '20 at 20:59