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The diagonal functor $\Delta_C^J:C \to C^J$ and the constant functors $\Delta_C^J(c):J\to C$ definitions are a bit too generous and lead to contradictions when applied to $J=0$ (the initial category). Let's see why.

According to the definitions, for every object $c$ in $C$ $\Delta_C^0(c)$ is the unique functor $0\to C$. This functor is indeed vacuously constant.

This creates a contradiction between - for example - the following two statements in MacLane'CWM, 2nd ed, when you set $J=0$.

1- Page 90 exercise 8a :

If the category $J$ is connected, $\lim(\Delta_C^J(c))=c$

Indeed: $0$ is vacuously connected and $\Delta_C^0(c_1)=\Delta_C^0(c_2)$, yet their limits are different ($c_1$ and $c_2$), for any two non isomorphic $c_1$ and $c_2$ objects.

2- Page 71

"a limit of the empty functor to $C$ is the terminal object of $C$".

So the functor $0\to C$ does not always have a limit, and when it does, it does not follow from the formula $\lim(\Delta_C^J(c))=c$ above.

The other extreme case ($C=0$) also lead to questionable definitions.

So I think the definition of the diagonal functor should be limited to non empty $J$ and $C$.

Question: am I missing something? or do you agree?

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The category $\mathbf{0}$ is not connected: – Oskar Jan 24 at 14:49
See also Similarly the empty space is not connected, the empty graph is not connected, $1$ is not prime, etc. – Qiaochu Yuan Jan 24 at 18:02
up vote 13 down vote accepted

It appears that (as Oskar mentions in the comments) the correct notion of "connected" for a category is usually a category which is inhabited (i.e. has at least one object) and such that between every pair of objects there is a zigzag of arrows connecting them. It's a small oversight on MacLane's part.

It's a recurring theme in category theory (and algebraic topology), one has to be careful about some definitions to make sure the empty case is handled properly. It's the same with topological spaces: a topological space is typically called "connected" if it has exactly two clopen subspaces, namely the empty set and itself. This prevents the space from being empty. Since it's reasonable to expect that a category is connected iff its nerve is, then this excludes the empty category.

Another example to understand why the empty space is often not said to be connected is the following reformulation: "a space $X$ is connected iff $\operatorname{Map}(X,-)$ preserves coproducts". In other words, maps from $X$ to a disjoint union $A \sqcup B$ are either maps $X \to A$ or maps $X \to B$. But $$\operatorname{Map}(\varnothing, A \sqcup B) = \{\varnothing\} \not\cong \operatorname{Map}(\varnothing,A) \sqcup \operatorname{Map}(\varnothing,B) = \{\varnothing_A, \varnothing_B\}.$$

(And in fact you see that my sentence that begins with "in other words" is a bit ambiguous: a map $\varnothing \to A \sqcup B$ is either a map $\varnothing \to A$ or a map $\varnothing \to B$, but both of these are the same...!)

The exact same thing happens with categories. You've also correctly identified one other reason that the empty category shouldn't be called connected: the limit of the empty diagram is indeed always the terminal object if it exists (this is correct), whereas if $\varnothing$ were connected the limit of the "constant" functor $\varnothing \to \mathcal{C}$ that maps "everything" to $c$ would be $c$.

PS: And in fact the identity $\varnothing \to \varnothing$ is not constant! For the (correct, IMO) definition of a constant map $X \to Y$ is $\exists y \in Y, \forall x \in X, f(x) = y$. This explains why the definition "$X$ is contractible iff $\operatorname{id}_X$ is homotopic to a constant map" is correct and excludes the empty space. (Thanks to Jeremy Rickard for pointing out an error in an earlier version of this paragraph.)

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I agree with you about "connected", but I think you're wrong about the definition of "constant". Your point about $\emptyset$ not being contractible only requires that the map $\emptyset\to\emptyset$ is not constant, and I agree that it's not. But by also requiring that $\emptyset\to Y$ is not constant for non-empty $Y$, you break all sorts of other things, such as the fact that every map to a contractible space is homotopic to a constant map, or that restrictions of constant functions are constant. – Jeremy Rickard Jan 25 at 9:11
The correct definition of a constant map $f:X\to Y$ is $\exists y\in Y,\forall x\in X,f(x)=y$. Or in other words, $f$ factors through the terminal object in the category of sets (a singleton). Similarly a functor is constant if it factors through the terminal category. So $\emptyset\to Y$ is constant iff $Y$ is non-empty. – Jeremy Rickard Jan 25 at 9:14
@JeremyRickard I think you're right, sorry for the oversight on my part. The empty set is tricky! Thanks for the remark. – Najib Idrissi Jan 25 at 9:18
Thank you @JeremyRickard and Najib for the enlightening discussion – magma Jan 26 at 17:58

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