Forgetful functor from the slice category creates limits

category-theoryfunctorslimits-colimits

I am reading "Category theory in context" and I'm having difficulties with proposition 3.3.8, top of page 92, pdf.

Theorem: The forgetful functor $F : c/C \to C$ strictly creates all limits that $C$ admits.

Proof (paraphrased): Defining a diagram over a slice category $J \to c/C$ is equivalent to defining a functor $K : J \to C$ and a cone $\kappa : \Delta c \Rightarrow K$. Suppose $K$ has a limit cone $\lambda: \Delta l \Rightarrow K$, then we'll show that this cone lifts to a cone over the slice category.

Since $\lambda$ is a limit cone in $C$, there's a unique factorisation of the cone $\kappa : \Delta c \Rightarrow K$ through $\lambda$, call it $t : c \to l$.

The book says that it's straightforward to verify that $t$ is a limit for the diagram in $c/C$.

Clearly $t$ is a summit of the cone. I can take the $i$-th leg to be $\lambda_i$, which is a morphism in the slice category from $t$ to $\kappa_i$.

It remains to show this cone is a limit cone. If I have a cone with a summit $v : c \to m$ and legs $\mu_i : m \to K(i)$, then I have to show there's a unique morphism in the slice category from $v$ to $t$ that commutes with the legs. I see only one possible candidate: because $\lambda$ is a limit in $C$, there's a morphism $p : m \to l$ in $C$ that commutes with the legs. But why is $p$ a morphism in the slice category, i.e. why does $p \circ v = t$?

Best Answer

The equality $p\circ v=t$ follows from the fact that the $\lambda_i$ for $i\in J$, being a limit cone, are jointly monomorphic (also said to be a monosource) and $$\lambda_i\circ p\circ v=\mu_i\circ v=\kappa_i=\lambda_i\circ t$$ for every $i$.

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