On page 39 of Category Theory for Working Mathematicians, Mac Lane makes a claim that seems to me to be false.

Let $\mathcal{C}$ be a category, and let $\mathbf{2}$ be the category with exactly two distinct objects $a$ and $b$ and exactly one non-identity arrow $\uparrow$ from $a$ to $b$. Let $\mu$ be the function that assigns to each object $(x,\delta)\in\mathcal{C}\times\mathbf{2}$ the arrow $({1}_{x},\uparrow)$. Mac Lane calls $\mu$ the universal natural transformation from $\mathcal{C}$.

Here's the claim: for all categories $\mathcal{D}$ and all functors $S,S':\mathcal{C}\rightarrow\mathcal{D}$ and natural transformations $\tau:S\rightarrow S'$, there is a unique functor $F:\mathcal{C}\times\mathbf{2}\rightarrow\mathcal{D}$ such that for all objects $(x,\delta)\in\mathcal{C}\times\mathbf{2}$, we have $$(*)\qquad F(\mu(x,\delta))=\tau x.$$ My doubt is about the uniqueness of such a functor. For a possible counterexample to uniqueness, take $\mathcal{C}$ to be the category with one object $a$ and two arrows ${1}_{a}$ and $h$, with $h\circ h:={1}_{a}$. Take $\mathcal{D}$ to be $\mathcal{C}\times\mathbf{2}$. Let $S:\mathcal{C}\rightarrow\mathcal{D}$ be the unique functor that sends $a$ to $(a,a)$ and $h$ to $(h,{1}_{a})$. Let $S':\mathcal{C}\rightarrow\mathcal{D}$ be the unique functor that sends $a$ to $(a,b)$ and $h$ to $(h,{1}_{b})$. Let $\tau:S\rightarrow S'$ be the natural transformation that sends $a$ to the arrow $({1}_{a},\uparrow)$. Let $F:\mathcal{C}\times\mathbf{2}\rightarrow\mathcal{D}$ be the identity functor. Then clear $F$ has property $(*)$. But now let $F':\mathcal{C}\times\mathbf{2}\rightarrow\mathcal{D}$ send each object to itself, send each arrow of the form $({1}_{a},g)$ to itself, and send each arrow of the form $(h,g)$ to $({1}_{a},g)$. Then $F'$ is another functor, distinct from $F$, with property $(*)$.

I've checked the counterexample several times and can't find any errors. Am I missing something?


1 Answer 1


No, your point is right. It's a bit mysterious why Mac Lane writes $F\mu c=\tau c$, as clearly the needed condition is $F\mu=\tau$, a condition which implies in particular that $FT_0=S$ and $FT_1=T$, using Mac Lane's notation, at least in the first edition. Mac Lane even writes the correct condition at the end of the paragraph.

Note that you reproduced some of the construction slightly incorrectly. $\mu$ is the function that assigns $(x,\uparrow)$ to each $x\in \mathcal C$, not to each $(x,\delta)$, because $\mu$ is a natural transformation with domain $\mathcal C$. Thus the claim you starred should also have been that $F(\mu(x))=\tau x$, not $F(\mu(x,\delta))=\tau x$.

  • 1
    $\begingroup$ Ah yes, thanks for pointing out my own mistake. Could you clarify why it's mysterious that Mac Lane writes that F composed with mu agrees with tau on objects? Isn't this the most we can say, seeing as tau is entirely determined by where it sends the objects of C? $\endgroup$
    – Anonymous
    Sep 25, 2018 at 1:37
  • $\begingroup$ @GregoryCarroll A natural transformation between two given functors is determined by its action on objects, but a functor out of $C\times \mathbf{2}$ contains more information than that, specifically the two functors the given natural transformation maps between. $\endgroup$ Sep 25, 2018 at 2:45
  • $\begingroup$ @KevinArlin I thought that $F\mu$ was defined as $(F\mu)_c=F\mu_c$ ($=\tau_c$ in this case) for all $c$ in $\mathcal{C}$, which doesn't seem to provide any additional information about $F$. It is enough to ensure that $FT_{0}$ agrees with $S$ and $FT_{1}$ with $T$ on objects, but I don't see why they should also agree on arrows (at least not all of them). $\endgroup$ Aug 17, 2021 at 20:14
  • $\begingroup$ @KevinArlin I mean, regardless of the meaning of the notation $F\mu=\tau$, what if there were pair of functors $S'$ and $T'$ such that $\tau:S'\longrightarrow T'$ is a natural transformation? Wouldn't that mean $F$ is not unique? $\endgroup$ Aug 17, 2021 at 20:35
  • $\begingroup$ @ModestoRosado A functor $C\times \mathbf 2\to D$ determines two functors $C\to D$ by simply restricting along the two obvious embeddings of $C$ into $C\times\mathbf 2.$ $\endgroup$ Aug 17, 2021 at 21:48

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