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Stream: learning: questions

Topic: Sheafification, adjunction between bundles and presheaves

Peva Blanchard (May 17 2024 at 22:44):

Another side quest from @David Egolf 's thread on reading Baez topos series.

Let $X$ be an arbitrary topological space.

First, we defined a functor $\Lambda$ from the category $\text{Psh}(X)$ of presheaves over $X$ to the category $\text{Top}/X$ of bundles over $X$. $\Lambda$ maps a presheaf $F$ over $X$ to an étale space $\Lambda(F)$ over $X$.

Second, we defined a functor $\Gamma : \text{Top}/X \rightarrow \text{Psh}(X)$. $\Gamma$ maps a bundle $E \rightarrow X$ to its sheaf of sections over $X$.

I was wondering if there were an adjunction $\Lambda \dashv \Gamma$.

I managed to get a proof that, indeed, $\Lambda$ is left adjoint to $\Gamma$. But my proof is relatively long, and I am not sure I checked everything right. An external confirmation/refutation is welcome.

ps: According to John in the other thread, the composite $\Gamma \circ \Lambda$ is called "sheafification", and appears as the left adjoint to the forgetful functor from sheaves to presheaves. This led me to wonder if sheafification were a monad.

Eric M Downes (May 17 2024 at 22:56):

Peva Blanchard said:

I managed to get a proof that, indeed, $\Lambda$ is left adjoint to $\Gamma$. But my proof is relatively long, and I am not sure I checked everything right. An external confirmation/refutation is welcome.

Was your proof too long to fit in the margins of this stream? :P

(Seriously, where do I find it? I'd like to read it.)

John Baez (May 17 2024 at 23:13):

Peva Blanchard said:

I was wondering if there were an adjunction $\Lambda \dashv \Gamma$.

Try the course notes: the end of Part 2 and also the section "The adjunction between presheaves and bundles" in Part 3.

John Baez (May 17 2024 at 23:16):

This led me to wonder if sheafification were a monad.

Since it's a left adjoint followed by a right adjoint, it's a monad. But even better, I believe it's an [[idempotent monad]]. That is, if you sheafify a presheaf that's already a sheaf, you don't get anything new, just something isomorphic to what you had.

Peva Blanchard (May 18 2024 at 05:29):

John Baez said:

Peva Blanchard said:

I was wondering if there were an adjunction $\Lambda \dashv \Gamma$.

Try the course notes: the end of Part 2 and also the section "The adjunction between presheaves and bundles" in Part 3.

Oops ! I should have known the answer was already there. I read the series last year: perhaps I was just remembering the adjunction without realizing that I read about it there.

Anyway, thank you for the link.

Peva Blanchard (May 18 2024 at 05:35):

Eric M Downes said:

Peva Blanchard said:

I managed to get a proof that, indeed, $\Lambda$ is left adjoint to $\Gamma$. But my proof is relatively long, and I am not sure I checked everything right. An external confirmation/refutation is welcome.

Was your proof too long to fit in the margins of this stream? :P

(Seriously, where do I find it? I'd like to read it.)

Yes sure :)

I chose to prove the adjunction by building a natural isomorphism

$$(Top/X)(\Lambda(F), E) \cong Psh(F, \Gamma(E))$$

where $F$ is a presheaf over $X$, and $\pi : E \rightarrow X$ a bundle over $X$.

Some notations. Given a section $s \in FU$ over an open subset $U \subseteq X$, $[s]_x$ is the germ of $s$ at $x$, and $[s] : U \rightarrow \Lambda(F)$ is the continuous function $x \mapsto [s]_x$. We also denote by $[U, s] = \{[s]_x ~|~ x \in U \}$ the set of germs of $s$ over $U$, which is a member of a base of open sets of $\Lambda(F)$.

Forward direction

Consider a bundle morphism $f : \Lambda(F) \rightarrow E$, i.e., a continuous function that preserves the base point. We want to define a natural transformation $\eta$ from the presheaf $F$ to the sheaf $\Gamma(E)$ of sections of $E$.

Components

Let $U$ be an open subset of $X$, and $s \in FU$ a section of $F$ over $U$.
We want to define a section $z$ of $E$ over $U$, i.e., a continuous function $z : U \rightarrow E$ such that $\pi\circ z = id_U$. For every $x \in U$, let's define

$$z(x) = f([s]_x) \in E$$

Because $f$ is a morphism of bundles over $X$, we have $\pi \circ z = id_U$.
It remains to show that $z$ is continuous. By construction of $\Lambda$, we know that the map

$$\begin{align*} [s] : U &\rightarrow \Lambda(F) \\ x &\mapsto [s]_x \\ \end{align*}$$

is continuous. By the continuity of $f$ and the fact that the composition of two continuous functions is continuous, we conclude that $z$ is also continuous.

To sum up, we defined a function

$$\begin{align*} \eta_U : FU &\rightarrow \Gamma(E)U \\ s &\mapsto f \circ [s] \end{align*}$$

Naturality

Consider an inclusion $U \subseteq V$. We want to show that the following diagram commutes.
image.png

This amounts to show that

$$\begin{align*} f \circ [s|_U] &= f \circ [s]|_U \\ &= \left(f \circ [s]\right)|_U \end{align*}$$

Therefore, we have a natural transformation $\eta : F \rightarrow \Gamma(E)$.

Peva Blanchard (May 18 2024 at 05:51):

Backward direction

Let $\eta : F \rightarrow \Gamma(E)$ be a natural transformation. We want to define a morphism $f : \Lambda(F) \rightarrow E$ of bundles, i.e. a continuous function that preserves the base point.

Definition

Fix a germ $[s]_x \in \Lambda(F)$, for some section $s \in FU$ over an open neighborhood $U \subseteq X$ of $x$. We define

$$f([s]_x) = [\eta_U s]_x$$

We need to show that $f$ is well-defined. Assume $[s]_x = [t]_x$ for another section $t \in FV$ over another open neighborhood $V \subseteq X$ of $x$. Without loss of generality, we may assume that $V \subseteq U$. In that case $t = s|_V$. Then

$$\begin{align*} [\eta_V t]_x &= [\eta_V s|_V]_x \\ &= [(\eta_U s)|_V]_x \text{ (naturality)}\\ &= [\eta_U s]_x \\ \end{align*}$$

Therefore, $f$ is well-defined. Note also that, by definition, $f$ preserves the base point.

Continuity

Let $[W, r]$ be an open neighborhood of $f([s]_x)$ in $\Gamma(E)$. Without loss of generality, we may, again, assume that $W \subseteq U$. Then,

$$r = \left(\eta_U s\right)|_W = \eta_W s|_W$$

is the restriction of $\eta_U s$ on $W$.

Consider $t = s|_W \in FW$, and the open neighborhood $[W, t]$ of $[t]_x = [s]_x$. For every $y \in W$,

$$\begin{align*} f([t]_y) &= [\eta_W t]_y \\ &= [\eta_W s|_W]_y \\ &= [\left(\eta_U \circ s\right)|_W]_y \\ &= [r]_y \\ &\in [W, r] \end{align*}$$

In other words,

$$[W, t] \subseteq f^{-1}[W, r]$$

I.e., $f$ is continuous.

Peva Blanchard (May 18 2024 at 05:51):

Natural bijection

We have shown that

$$\begin{align*} (Top/X)(\Lambda(F), E) &\xrightarrow{A} Psh(F, \Gamma(E)) \\ f &\mapsto \eta = \left( s \xmapsto{\eta_U} f \circ [s] \right)_{U \in \mathcal{O}X} \end{align*}$$

and

$$\begin{align*} Psh(F, \Gamma(E)) &\xrightarrow{B} (Top/X)(\Lambda(F), E) \\ \eta &\mapsto \left([s]_x \text{ with } s \in FU \xmapsto{f} [\eta_U s]_x \right) \end{align*}$$

are both well-defined functions.

Prove that $B \circ A = id$

Let $f : \Lambda(F) \rightarrow E$ be a morphism of bundles over $X$, and $g = B(A(f))$.
By definition, for every $s \in FU$ for some open neighborhood $U$ of $x$,

$$\begin{align*} g([s]_x) &= [A(f)_U s]_x \\ &= [f \circ [s]]_x \\ &= f([s]_x) \end{align*}$$

I.e., $g = f$, and thus $B \circ A = id$.

Prove that $A \circ B = id$

Let $\eta : F \rightarrow \Gamma(E)$ be a natural transformation of presheaves, and $\nu = A(B(\eta))$.
By definition, for every open subset $U$ of $X$,

$$\begin{align*} \nu_U : FU \rightarrow &~ \Gamma(E)U \\ s \mapsto &~ B(\eta) \circ [s] \\ &= \left(x \mapsto B(\eta)([s]_x) \right)\\ &= \left(x \mapsto [\eta_U s]_x \right)\\ &= \eta_U s \end{align*}$$

I.e., $\nu_U = \eta_U$, and thus $A \circ B = id$.

Peva Blanchard (May 18 2024 at 08:00):

Naturality of $A$

With respect to $E$

We show that $A = A_{F, E}$ transforms naturally with respect to $E$.

Consider a morphism of bundles $g : E \rightarrow E'$.
The naturality condition is expressed in the following diagram

image.png

I.e., we want to show that, for every open subset $U \subseteq X$,

$$\Gamma(g)_U \circ \eta_U = \eta'_U$$

I.e., for every $s \in FU$,

$$\begin{align*} &\Gamma(g)_U(\eta_U(s)) = \eta'_U(s) \\ \iff &\Gamma(g)_U(f \circ [s]) = (g \circ f) \circ [s] \\ \iff &g \circ (f \circ [s]) = (g \circ f) \circ [s] \end{align*}$$

which holds by the associativity.

With respect to $F$

We show that $A = A_{F, E}$ transforms naturally with respect to $F$.

Consider a natural transformation $\lambda : F \rightarrow F'$ of presheaves over $X$. The naturality condition is expressed in the following diagram
image.png

I.e., we want to show that, for every open subset $U \subseteq X$,

$$\eta'_U = \eta_U \circ \lambda$$

I.e., for every section $s \in F'U$,

$$\begin{align*} &\eta'_U(s) = \eta_U(\lambda_U(s)) \\ \iff &(f \circ \Lambda(\lambda))(s) = f \circ [\lambda_U(s)] \\ \iff &f \circ (\Lambda(\lambda)(s)) = f \circ [\lambda_U(s)] \\ \iff &f \circ [\lambda_U(s)] = f \circ [\lambda_U(s)] \\ \end{align*}$$

Peva Blanchard (May 18 2024 at 08:47):

Naturality of $B$

We have shown that every component $B_{F,E}$ is the inverse of the component $A_{F,E}$ (as functions in $Set$). Because $A$ is natural in $E$ and $F$, I think we can conclude that $B$ is also natural in $E$ and $F$.

Peva Blanchard (May 18 2024 at 09:10):

This concludes the proof that $\Lambda$ is left adjoint to $\Gamma$.

John Baez (May 18 2024 at 09:13):

Wow, great! When @David Egolf reaches this material in the course, he can either try to figure out himself why $\Lambda$ is left adjoint to $\Gamma$ and check his work here, or just read this!