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Consider a category $\mathsf C$ admitting a quadruple adjunction as below.

$$(\Pi_0 \dashv \text{disc} \dashv \Gamma \dashv \text{codisc}) : \mathsf{C} \stackrel{\stackrel{\longrightarrow}{\longleftarrow}}{\stackrel{{\longrightarrow}}{{\longleftarrow}}} \mathsf{B} \;$$ A good example is the category of locally connected spaces with $\mathsf{B}=\mathsf{Set}$. In this case the connected-components functor $\Pi_0$ implies $\mathsf C$ is extensive.

Suppose now $\mathsf C$ is extensive and admits the exponentials below. In Axiomatic Cohesion, Lawvere defines an adjoint quadruple as above to be cohesive if:

  1. $\Pi_0$ preserves finite products and $\text{codisc}$ is fully faithful.
  2. $\Pi_0$ preserves $\mathsf{B}$-parametrized powers in the sense of a natural isomorphism $$\Pi_0(X^{\text{disc}I})\cong \Pi_0(X)^{I}.$$
  3. The canonical arrow $\Gamma\Rightarrow \Pi_0$ is epimorphic.

Lawvere later defines such an adjunction to be sufficiently cohesive if each object has a "contractible envelope":

  1. For every $X$ there exists a monic map $X\to Y$ with $Y$ contractible in the sense that $\Pi_0(Y^A)=\mathbf 1$ for all $A$.

Then it is proved that a topos of cohesion ($\mathsf C$ is a topos) is sufficiently cohesive iff the truth-value object is connected, and also iff all injective objects are connected. Here, connectedness of $X$ means $\Pi_0(X)\cong \mathbf 1$.

For locally connected spaces it seems the injective objects are the inhabited indiscrete spaces, whose connectedness reflects their cohesion. This is opposed to case of sets where $\text{codisc}$ is the identity and sets are seldom singletons.

Can a sufficiently cohesive category (quadruple adjunction) have a conservative $\Gamma$?

I ask for the following reason: For locally connected spaces $\Gamma$ is the "points functor" and I think of the fact it is faithful but not conservative as reflecting that "discontinuous" arrows (i.e breaking cohesion) may be underlain by "isomorphisms on stuff"; I want to know if this is "correct" intuition or not.

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  • $\begingroup$ Just a self-note to start thinking about the answer: $\Gamma$ is conservative iff the counit of $d\dashv \Gamma$ is a strong epi. There are many categories on the $n$Lab page that exhibit cohesion over (s)Set or another topos, you can start checking how often $\Gamma$ is conservative with this criterion. $\endgroup$ – Fosco Nov 13 '17 at 0:44
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Let ${\phi : \mathrm{disc} \rightarrow \mathrm{codisc}}$ be the canonical map.

The map ${\Gamma \phi }$ is an iso in the base so, if $\Gamma$ is conservative, $\phi$ is an iso.

In other words, if $\Gamma$ is conservative, the original string of adjoints is a quality type. So, if the original string is sufficiently cohesive then the whole thing is degenerate.

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