I'm condusion on a statement in this page comma object in $n$lab. It states:

any strict comma object is a comma object, but the converse is not in general true.

My confusion is: the strict comma object trivially satisfies the $1$-dimensional univerality, but how to show it satisfies the $2$-dimensional one?


Your understanding of the definition is incorrect, but that is probably because the cited nLab page is misleading. Let me spell it out a little bit more accurately:

Let $\mathfrak{K}$ be a 2-category and let $f : A \to C$ and $g : B \to C$ be morphisms in $\mathfrak{K}$. A strict comma object $(f \downarrow g)$ in $\mathfrak{K}$ is an object $P$ in $\mathfrak{K}$ together with isomorphisms $$\mathfrak{K} (T, P) \cong (\mathfrak{K} (T, f) \downarrow \mathfrak{K} (T, g))$$ that are 2-natural in $T$.

In particular, strict comma objects have a 1-dimensional universal property as well as a 2-dimensional universal property by definition. The 1-dimensional universal property is as stated on nLab: so we have a (not necessarily commutative!) diagram in $\mathfrak{K}$ of the form below, $$\require{AMScd} \begin{CD} P @>{q}>> B \\ @V{p}VV @VV{g}V \\ A @>>{f}> C \end{CD}$$ and a 2-cell $\alpha : f \circ p \Rightarrow g \circ q$ such that etc. The 2-dimensional universal property says exactly this:

Given morphisms $x_0, x_1 : T \to P$ and 2-cells $\phi : f \circ p \circ x_0 \Rightarrow f \circ p \circ x_1$ and $\psi : g \circ q \circ x_0 \Rightarrow g \circ q \circ x_1$ such that $\psi \bullet \alpha x_0 = \alpha x_1 \bullet \phi$, there is a unique 2-cell $\theta : x_0 \Rightarrow x_1$ such that $\phi = f p \theta$ and $\psi = g q \theta$.

This, you will notice, is what appears on nLab. So the difference between strict comma objects and (bicategorical) comma objects is elsewhere – in fact, it is in the 1-dimensional universal property: one has to replace the uniqueness clause with an appropriate up-to-isomorphism version so that the resulting notion is invariant under equivalences in $\mathfrak{K}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.