Franklin's Notes

Universal arrow

If $S:\mathsf D\to\mathsf C$ is a functor and $c$ is an object of the category $\mathsf C$, a universal arrow from $c$ to $S$ is a pair $\langle r,u\rangle$ consisting of an object $r\in\mathsf D$ and an arrow $u:c\to Sr$ such that for every pair $\langle d,f\rangle$ with $d$ an object of $\mathsf D$ and $f:c\to Sd$ an arrow of $\mathsf C$, there exists a unique arrow $f':r\to d$ making the following diagram commute:

In other words, $u:c\to Sr$ is universal when the pair $\langle r,u\rangle$ is an initial object in the comma category $c\downarrow S$. If $\mathsf C=\mathsf{Set}$, then we say that a universal element of $S$ consists of a pair $\langle r,e\rangle$ where $r$ is an object of $\mathsf D$ and $e\in Sr$ so that for every pair $\langle d,x\rangle$ with $x\in Hd$, there is a unique arrow $f:r\to d$ with $Sf(e)=x$. This can be equivalently thought of as a universal $\mathsf{Set}$-arrow in which $c=1$ is the terminal object in $\mathsf{Set}$. In this way, we can generalize the idea of a "universal element" to any category with a terminal object.

Let's look at some examples, most of which are taken from MacLane.

If we let $\mathsf{Vect} \mathbb k$ be the category of vector spaces over the field $\mathbb k$, we may consider the forgetful functor $U:\mathsf{Vect} \mathbb k\to\mathsf{Set}$ which sends each vector space to its underlying set of elements. For any set $X$ we may form the vector space $V_X$ whose basis elements are the elements of $X$, so that the elements of $V_X$ are formal linear combinations of elements of $X$ in $\mathbb k$. If $j:X\to UV_X$ is the inclusion map, then $j$ is a universal arrow. For if $f:X\to UW$ is an arrow in $\mathsf{Set}$ for some $\mathbb k$-vector space $W$, we have that $f$ determines a linear transformation $f':V_X\to W$, for the image of any formal $\mathbb k$-linear combination of elements of $X$ can be mapped to the corresponding linear combination of the images of the basis vectors of $X$ in $W$.

Here's another example. As a consequence of the Yoneda lemma , the identity natural transformation $1:\mathsf C(c,-)\Rightarrow \mathsf C(c,-)$ is a universal arrow in the functor category $\mathsf{Set} ^\mathsf C$. In other words, for any natural transformation $\alpha:\mathsf C(c,-)\Rightarrow \mathsf C(d,-)$, there exists a unique morphism $f$ of $\mathsf C$ making the following diagram commute:

The property of a functor being representable can also be expressed as a universal property. In particular, a functor $K:\mathsf C\to\mathsf{Set}$ is representable iff there exists a universal arrow $u$ from the one-element set $1={\varnothing}$ to $K$:

In particular, if $u:1\to Kd$ is a universal arrow from $1$ to $K$, then we may construct a natural isomorphism $\alpha:\text{Hom}(d,-)\to K$ by letting the component $\alpha_x$ send each morphism $g\in\text{Hom}(d,x)$ to the element $Kg\circ u(\varnothing)$ in $Kx$. Naturality is straightforward to demonstrate using the functoriality of $K$.

Proposition 1. Let $F:\mathsf C\to\mathsf D$ be a functor with the property that for every object $y\in\mathsf D$, there exists a universal arrow $j:y\to Fx$ for some object $x$ of $\mathsf C$. Then there exists a functor $G:\mathsf D\to\mathsf C$ such that there is a universal arrow $j:y\to FGy$ for every $y\in\mathsf D$.

Proof. Let $F$ be a functor with the stated property. Define $G$ on objects by letting $Gy$ equal one of the objects $x\in\mathsf C$ such that there is a universal arrow $j:y\to Fx$, for each object $y$ of $\mathsf D$. Define $G$ on morphisms as follows: if $h:y\to y'$ is a morphism of $\mathsf D$, let $Gh$ be the unique morphism $h'$ of $\mathsf C$ such that $(Fh')j=j'h$, where $j:y\to FGy$ and $j':y'\to FGy'$ are universal arrows, so that the existence and uniqueness of $h'$ follow from universality.

It remains to show that $G$ is actually a functorial way of mapping arrows to arrows. To see why this is the case, consider the following diagram:

where $y_1,y_2,y_3$ are arbitrary objects of $\mathsf D$ with arrows $h_1:y_1\to y_2$ and $h_2:y_2\to y_3$, and universal arrows $j_1:y_1\to FGy_1$, $j_2:y_2\to FGy_2$, and $j_3:y_3\to FGy_3$. Recall that by universality of the $j_i$, the arrows $Gh_1$ and $Gh_2$ are uniquely chosen to make the two squares commute. This means that the outer rectangle must also commute. However, by the definition of $G$ and universality of $j_1$, we have that $G(h_2\circ h_1)$ is supposed to be the unique arrow $FGy_1\to FGy_3$ making the outer rectangle commute. This means that we must have $Gh_2\circ Gh_1=G(h_2\circ h_1)$, proving functoriality. $\blacksquare$


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