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In Hovey, Model Categories, he says that given a category $\mathcal{C}$ with a small limits and colimits, if we form the category of pointed objects ,i.e. the category with objects as morphisms $*\rightarrow A$ where * is the terminal object, then this category $\mathcal{C}_*$also has all small limits and colimits.

To prove this, for any Functor $F:J\rightarrow \mathcal{C}_* $ where $J$ is small we compose it with the forgetful functor $\mathcal{C}_* \rightarrow \mathcal{C}$ and take the limit here. My question is how will this give a limit in $\mathcal{C}_*$? we will get an object in the category, but how do we get the arrow from $*$ to that object?

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    $\begingroup$ You can use the universal property of the limit $\lim (\pi\circ J)$ of $J\to {\mathcal C}_{\ast}\stackrel{\pi}{\to}{\mathcal C}$ to construct a unique morphism ${\ast}\to \lim (\pi\circ J)$ such that all projections $\lim (\pi\circ J)\to J(j)$ for $j\in J$ are morphisms in ${\mathcal C}_{\ast}$ w.r.t. this map. $\endgroup$
    – Hanno
    Feb 17, 2015 at 10:21
  • $\begingroup$ Thanks, that works. Does this mean this statement will be true for any slice category of C? $\endgroup$
    – Arun Kumar
    Feb 17, 2015 at 10:30
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    $\begingroup$ Is true for slices under an object. Of course, the dual statement is true for slices over an object. $\endgroup$
    – Zhen Lin
    Feb 17, 2015 at 13:50

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Here is how it works in general: If $\mathcal{C}$ is a category and $P$ is an object (in your case, a final object), then the forgetful functor $P/\mathcal{C} \to \mathcal{C}$ creates limits. In particular, if $\mathcal{C}$ is complete, then $P/\mathcal{C}$ is complete, too. To see this, consider a diagram $(P \to X_i)$ in $P/\mathcal{C}$ and consider a limit cone $(\lim_i X_i \to X_i)$. Since the $(P \to X_i)$ can be regarded as a cone, it lifts to a morphism $P \to \lim_i X_i$. Then one verifies without effort that $((P \to \lim_i X_i) \to (P \to X_i))$ is a limit cone in $P/\mathcal{C}$.

The case of colimits is more complicated, since the forgetful functor doesn't create colimits. However, if $\mathcal{C}$ is cocomplete, then $P/\mathcal{C}$ will be cocomplete: An initial object is $(\mathrm{id} : P \to P)$. The coproduct of a non-empty family of objects $(P \to X_i)$ is given by the wide pushout $\coprod_P X_i$ in $\mathcal{C}$ equipped with the morphism $P \to X_i \to \coprod_P X_i$ for some $i$ (it doesn't matter which one). The coequalizer of two morphisms $P \to (X \rightrightarrows Y)$ is the coequalizer $Y \to C$ of the underlying morphisms in $X \rightrightarrows Y$ equipped with the morphism $P \to Y \to C$. Coequalizers are created by $P/\mathcal{C} \to \mathcal{C}$.

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