$f^+\mathscr{G}$ not a sheaf even if $\mathscr{G}$ is

Question

I have read that there are continuous maps $f:X\to Y$ and sheaves $\mathscr{G}$ on $Y$ such that $f^+\mathscr{G}$ is not a sheaf. Do you have an easy example for me?

Answer 1

The following counterexample should work, if I am not missing anything.

Let $X$ be an infinite set equipped with the cofinite topology (the one where closed subsets are given by the finite subsets and the whole $X$). Since $X$ is infinite, it is connected (if $U\subsetneq X$ is a proper, non-empty subset, it cannot be both open and closed, because it cannot be both infinite and finite). Furthermore, it is immediate to see that any subspace of a cofinite topological space is a cofinite topological space and, if this subspace is finite (as a set), then it is also discrete. In particular, our $X$ is a topological space such that every (non-empty) open subset is connected: this is the key point we need.

Indeed, let $Y$ be a finite subset of $X$ (having at least two distinct elements), endowed with the subspace topology and denote with $i\colon Y\to X$ the inclusion map. Let us consider a set $A$ which is not empty nor a singleton and the constant sheaf $\mathscr{F}$ over $X$ with fiber $A$ (so, up to isomorphisms, $\mathscr{F}$ is the sheaf of locally constant functions on open subsets of $X$, with values in $A$). Now, for each open subset $U\neq \emptyset$ of $X$, connectedness of $U$ means that $\mathscr{F}(U)\simeq A$. This implies that, given any non-empty subset $V$ of $Y$,

$$ (i^{+}\mathscr{F})(V)=\varinjlim_{V\subseteq U\subseteq X} \mathscr{F}(U)\simeq A. $$

So, for an open subset $V$ of $Y$, if $\mathscr{G}:=i^{+}\mathscr{F}$, $\mathscr{G}(V)\simeq \{0\}$, if $V=\emptyset$, whereas $\mathscr{G}(V)\simeq A$, if $V\neq\emptyset$. This is not a sheaf, because $Y$ admits disjoint open subsets $V,\ W$, with $\emptyset\neq V\neq W\neq\emptyset$ ($Y$ is a discrete space, so it is totally disconnected). Indeed, if $a\neq b$ are elements of $A$, $(a, b)$ is a compatible family for $\mathscr{G}$ with respect to $\{V,\ W\}$ (which is an open covering of $V\cup W$) and it clearly does not admit any amalgamation in $\mathscr{G}(V\cup W)\simeq A$.

---

EDIT: As required by the OP, I try to explain better why $\mathscr{G}(\emptyset)\simeq \{0\}$ above. Firstly, let us take a look at the definition of $(i^{+}\mathscr{F})(V)=\mathscr{G}(V)$; it might be worth saying that writing

$$\varinjlim_{V\subseteq U\subseteq X} \mathscr{F}(U)$$

is a sloppy (but quite acceptable, I would say) notation to mean the following. For $V\subseteq Y$ open, let $I_{V}$ be the full subcategory of $(\text{Open}(X))^{op}$ given by those $U\in\text{Open}(X)$ such that $V\subseteq U$ (here $\text{Open}(X)$ is the poset of open subsets of $X$, ordered by inclusion and seen as a category in the canonical way). If $\mathscr{F}_{\vert I_{V}}$ denotes the restriction of $\mathscr{F}$ to $I_{V}$, then

$$\varinjlim_{V\subseteq U\subseteq X} \mathscr{F}(U):=\varinjlim \mathscr{F}_{\vert I_{V}}.$$

Now, if $V$ is open also in $X$ (and of course this is the case when $V=\emptyset$), then $I_{V}$ has a terminal object, given by $V$. This implies that $\varinjlim \mathscr{F}_{\vert I_{V}}\simeq \mathscr{F}(V)$ thanks to an easy exercise in general category theory: given a functor $\mathscr{T}\colon\mathcal{J}\to\mathcal{C}$ between arbitrary categories, if $\mathcal{J}$ has a terminal object $1$, then $\varinjlim\mathscr{T}\simeq\mathscr{T}(1)$.

Answer 2

Let $X$ be a topological space containing two closed points $x,y$ and let $i : \{x,y\} \to X$ denote the inclusion map. Notice that $\{x,y\}$ carries the discrete topology. Let $F$ be a sheaf on $X$. Then $i^+ F$ is a presheaf on $\{x,y\}$ which is given by $(i^+ F)(\emptyset)=1$ (the terminal set), $(i^+ F)(\{x\}) = F_x$ (the stalk at $x$), $(i^+ F)(\{y\})=F_y$ (the stalk at $y$) and $$(i^+ F)(\{x,y\}) = \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U).$$ Hence, $i^+ F$ is a sheaf if and only if the canonical map $$\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y.$$ is an isomorphism. But this doesn't have to the case, even if $F$ is a sheaf. At least, the map is injective. When $x,y$ can be separated by disjoint open neighborhoods, then the map is surjective.

If $F$ is a sheaf of rings, then surjectivity of the map would imply that $(1,0)$ lies in the image, which means that there is a section $s \in F(U)$ with $x,y \in U$ such that $s_x = 1$ and $s_y = 0$. Now consider for example an integral scheme $X$ and its structure sheaf $F = \mathcal{O}_X$. Then $F(U) \to F_x$ is injective. Then $s_x=1$ implies $s=1$, but $s_y=0$ implies $s=0$, a contradiction. Hence, any integral scheme with two closed points (for example $\mathrm{Spec}(\mathbb{Z})$) provides a counterexample. This is just what came into my mind, I am sure that there are many other (possibly more natural) counterexamples.