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?
Best Answer
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 $F(\mu(x,\delta))=\tau x$.