The Things That Keep Me Up At Night

Niven Achenjang bio photo By Niven Achenjang Comment

Every now and then, I will find some theorem or exercise that is interesting and “obviously” doable, but despite my best efforts, I just can’t seem to figure out what is going on. Sometimes when this happens, I find myself unable to let go of this problem, and it enters my mathematical backlog, where it will continue to torment me until the day I die1. When I have free headspace, I like to return to these problems in hopes of finally having the insight I need; this post will be my attempt 2 to resolve one of the things that more recently kept me up at night. In particular, we will show that3

where $A$ is an abelian group and $A_{\R}$ is its associated constant sheaf on $\R$. A lot 4 of me figuring out the details of showing this will be done as I write this post, so what follows while likely be rough and ill-organized.

What Could Have Been

In the end, this post turned out very differently from what I originally imagined. The original game plan was, roughly, introduce the sheaf inverse image and show that given a (continuous) function $f:X\to Y$ and a sheaf $\ms F$ on $Y$, one gets maps on cohomology

After this, remark that the inverse image of a constant sheaf is a constant sheaf (!), so sheaf cohomology with constant coefficients is a (contravariant) functor. Finally, I would show that any two homotopic maps $f\simeq g:X\to Y$ induced the same map on cohomology

where $A$ is an abelian group and $A_Y$ is its associated constant sheaf on $Y$. This would mean that homotopy equivalent spaces have the same (sheaf) cohomology (with constant coefficients). Since $\R$ is contractible and any sheaf on a one-point space is flasque, this would prove that $\hom^1(\R,A_{\R})=0$.

However, while trying to figure out how to actually do all of that, I stumbled across this stackexchange question, which presents a much better 5 proof of this fact. Hence, I will instead follow the lead of that question. 6

I never did finish working out the details of my original plan, so it’s possible I return to it in the future. While I do not think it is the best method to achieve the aim of this post, I do think it would be nice to write up the details of why cohomology with constant coefficients is invariant under homotopy.

The Proof

One of the big things about cohomology is that short exact sequences (of sheaves) give rise to long exact sequences in cohomology. Because of that, the claim that $\hom^1(\R, A_{\R})=0$ is equivalent to the following.

Let $0\to A_{\R}\xrightarrow f\ms B\xrightarrow g\ms C\to0$ be a short exact sequence of sheaves (of abelian groups) on $\R$. Then, $0\to A\to\ms B(\R)\to\ms C(\R)\to0$ is exact. In particular, $\ms B(\R)\to\ms C(\R)$ is surjective.
Fix any $\gamma\in\ms C(\R)$, and let $$\mc S=\brackets{\left.(U,\beta)\right|g_U(\beta)=\gamma\mid_U\text{ and }U\text{ is an open interval}}\text{ where }(U,\beta)\le(U',\beta')\iff U\subseteq U'\text{ and }\beta'\mid_U=\beta$$ Now, consider some chain $(U_1,\beta_1)\le(U_2,\beta_2)\le\dots$ in $\mc S$. Let $U=\bigcup_nU_n$ and let $\beta="\lim\beta_n"$. By $\lim\beta_n$ what I really mean is that $\{U_n\}$ gives an open cover of $U$ and because $\ms B$ is a sheaf, the various $\beta_n\in U_n$ patch together to form a unique element $\beta="\lim\beta_n"$ of $U$. Note that $U$ is an open interval and that $g_U(\beta)=\gamma\mid_U$ essentially because $\ms C$ is a sheaf. Thus, every chain in $\mc S$ has an upper bound, so Zorn's lemma implies that $\mc S$ has a maximal element $(V,\delta)$. We claim that $V=\R$, so $\delta\in\ms B(\R)$ is a preimage of $\gamma$. Suppose otherwise. By exactness at $\ms C$, there exists an open cover $\{U_n\}$ of $\R$ with elements $\beta_n\in U_n$ s.t. $g_{U_n}(\beta_n)=\gamma\mid_{U_n}$ for all $n$. By decomposing each $U_n$ into its connected components, and replacing the corresponding $\beta_n$ with its restrictions onto those components, we may assume that each $U_n$ is an open interval. Now, since $V\neq\R$, there must exist some $U_m$ s.t. $U_m\not\subset V$ and $U_m\cap V\neq\emptyset$. Let $V_m=U_m\cap V$, and note that $$\delta\mid_{V_m}-\beta_m\mid_{V_m}\in\ker g=\im f$$ so there's some $\alpha_m\in A_{\R}(V_m)$ s.t. $f_{V_m}(\alpha_m)=\delta_{V_m}-\beta_m\mid_{V_m}$. Note that $V_m$ is an open interval, so $A_{\R}(V_m)=A=A_{\R}(\R)$ and the restriction map $A_{\R}(\R)\to A_{\R}(V_m)$ is the identity. Hence, $\alpha_m$ extends to a global section $\alpha\in A_{\R}(\R)$. Finally, $f_{U_m}(\alpha\mid_{U_m})+\beta_m\in\ms B(U_m)$ and $\delta\in\ms B(V)$ agree on $V_m$, so they patch together to form a (unique) $\beta\in\ms B(U_m\cup V)$. However, $$g_{U_m\cup V}(\beta)\mid_{U_m}=g_{U_m}(\beta_m-f_{U_m}(\alpha\mid_{U_m}))=\gamma\mid_{U_m}$$ and $$g_{U_m\cup V}(\beta)\mid_V=g_V(\delta)=\gamma\mid_V$$ so $g_{U_m\cup V}(\beta)=\gamma_{U_m\cup V}$. However, because $U_m\cup V\supsetneq V$, this contradicts $(V,\delta)$ being a maximal element. Thus, we must have had $V=\R$ all along, proving the theorem.
$\hom^1(\R,A_{\R})=0$
Just embed $A_{\R}$ in any acyclic sheaf. i.e. consider any short exact sequence $$0\to A_{\R}\to\ms B\to\ms C\to0$$ where $\ms B$ is acyclic. Then, looking at the long exact sequence on cohomology, we see $$0\to\hom^0(\R, A_{\R})\to\hom^0(\R, \ms B)\to\hom^0(\R, \ms C)\to\hom^1(\R, A_{\R})\to0$$ but $\im(\hom^0(\R,\ms B)\to\hom^0(\R, \ms C))=\hom^0(\R, \ms C)$ by the above theorem. By exactness, this means that $\ker(\hom^1(\R, A_{\R})\to0)=0$ so $\hom^1(\R, A_{\R})=0$.

A Second Question

In the previous section we showed that all constant sheaves on $\R$ are acyclic 7. One can similarly show that, for any fixed $k\ge0$, the sheaf $C^k_{\R,\R}$ of $k$-differentiable functions (where 0-differentiable=continuous) into $\R$ is acyclic. In fact, when you think about, sheaf cohomology to vaguely about measuring the failure of certain local-to-global problems on your space, but $\R$ looks very similarly locally and globally. In fact, given any point in $\R$, you can find arbitrarily small neighborhoods of that point that are homeomorphic to the whole space 8. Thinking about this made me wonder the following:

Are there any non-acyclic (cyclic?) sheaves (of abelian groups) on $\R$?

I brought this question up to one of my friends, and after talking about it for a while, we found an example 9 which shows the answer is yes. Quick note: for arbitrary $S\subset X$ and a sheaf $\ms F$ on $X$, we set

where the directed limit is taken over open sets $U\supseteq S$. Hence, elements of $\ms F(S)$ are “germs over $S$.”

Fix any two distinct points $a,b\in\R$, and let $U=\R\sm\brackets{a,b}$. Let $\Z_U$ be the sheaf on $\R$ given by the sheafification of10

where the $\Z_U$ in the definition above is the constant sheaf on $U$ and not the sheaf on $\R$ we are constructing. Let $\Z_{\{a,b\}}$ be a similarly defined sheaf on $\R$ with ${a,b}$ in place of $U$. Then, we get a short exact sequence

where exactness can easily be checked on stalks: for stalks over $x\not\in\{a,b\}$ the sequence becomes $0\to\Z\to\Z\to0\to0$ which is exact, and for stalks over $x\in\{a,b\}$ it becomes $0\to0\to\Z\to\Z\to0$ which is also exact. Looking at cohomology, we get that the following is exact

This means that $\hom^1(\R,\Z_U)\neq0$ as there’s no surjection $\Z\to\Z\oplus\Z$. Thus, $\Z_U$ is a non-acyclic sheaf on $\R$.

  1. or I figure it out; whichever happens first. 

  2. if you are reading this, then I probably succeeded. 

  3. Moreso than usual, this post is primarly meant for me, so many things won’t be defined as I assume I will still know them whenever I look back on this post in the future. 

  4. Most (I’ve really only thought about big picture stuff) 

  5. better in terms of simplicity. As a consequence though, it involves less general concepts/theorems. 

  6. I’ll basically just fill in the details of the first bullet point of the answer given in the link 

  7. This isn’t quite true, but it is not hard to show this after what was done in the previous section. 

  8. This is kind of the idea behind interval-flasque sheaves on R 

  9. in Hartshorne 

  10. I must admit that it is possible I am defining things incorrectly here. If you notice a mistake, call me out on it. 

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