Germ spaces, étalé maps and geometric sheafification

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A sheaf records sections on open sets; a local homeomorphism records how their germs fit together as a topological space. The topology is essential. A collection of stalks without that topology does not determine which families are sections, and a local homeomorphism need not be a covering map.

Throughout, $X$ is an arbitrary topological space. No separation, connectedness or path hypothesis is imposed. All sets and covers belong to fixed universes. Sheaves have values in sets and satisfy the empty-cover axiom, so their value on the empty open is a singleton. Maps of spaces over $X$ are continuous maps commuting with the projections. Write $\operatorname{Top}/X$ for that category and $\operatorname{Et}(X)$ for its full subcategory of local homeomorphisms. Here a local homeomorphism need not be surjective.

The point-free construction for general coefficient categories is in Associated sheaves with general coefficients, Sections 2–4. The present construction supplies a concrete topological model for set-valued sheaves. Locally constant module sheaves treats its separate module structure and transition maps.

1. The topology on the germs of a presheaf

Let $P$ be a presheaf. A germ at $x$ is an equivalence class $[s]_x$, where $s\in P(U)$ and $x\in U$. Two representatives are equivalent when their restrictions agree on some open neighborhood of $x$ contained in both domains. Reflexivity and symmetry follow directly; intersecting two such neighborhoods proves transitivity. Restricting representatives defines the same class. Put

\begin{gathered} E(P)=\coprod_{x\in X}P_x,\\ p_P([s]_x)=x. \end{gathered} \tag{1.1}

For each $s\in P(U)$, define a map and its image by $\begin{gathered} \widehat s:U\longrightarrow E(P),\\ x\longmapsto[s]_x,\\ [s,U]=\widehat s(U). \end{gathered} \tag{1.2}$

Proposition 1.1. The sets $[s,U]$ form a basis of a topology on $E(P)$. Each restriction $p_P:[s,U]\to U$ is a homeomorphism, and $p_P$ is a local homeomorphism.

Proof. Every germ has a representative, so these sets cover. If a germ lies in $[s,U]\cap[t,V]$, the equality of its representatives supplies an open neighborhood $W\subset U\cap V$ on which $s|_W=t|_W$. The basic set $[s|_W,W]$ contains that germ and is contained in the intersection. This proves the basis axiom, including the case in which the intersection is empty.

The projection restricts bijectively on $[s,U]$, with inverse $\widehat s$. The inverse image under $\widehat s$ of any basic set $[t,V]$ is the set of points in $U\cap V$ where the two germs agree. This is open, since each such equality holds on a smaller neighborhood. Thus $\widehat s$ is continuous. Also $p_P^{-1}(V)$ is the union of the basic sets $[s|_{U\cap V},U\cap V]$, so $p_P$ is continuous. Every open subset of $[s,U]$ is a union of intersections with basic sets; its projection is open by the same neighborhood calculation. Hence the displayed bijection is a homeomorphism. These charts cover $E(P)$. $\square$

A presheaf map $b:P\to Q$ induces $E(b)([s]_x)=[b_U(s)]_x. \tag{1.3}$

Naturality of $b$ makes this independent of representatives. On each chart it is the continuous map $\widehat{b_U(s)}\circ p_P$, so it is continuous on their open cover. Identities and compositions are respected at every germ. This defines a functor $E:\operatorname{PSh}(X)\to\operatorname{Et}(X)$, and also a functor into $\operatorname{Top}/X$.

2. Sections recover the entire étalé space

For any continuous $q:T\to X$, let $\Gamma_X(T)(U)$ be the continuous maps $t:U\to T$ satisfying $qt=1_U$. Restrictions restrict maps. Compatible sections glue uniquely as functions; the glued function is continuous because continuity can be checked on an open cover. Thus $\Gamma_X(T)$ is a sheaf, even when $q$ is not a local homeomorphism. Composition with a map over $X$ makes $\Gamma_X$ functorial.

Lemma 2.1. If $q$ is a local homeomorphism, every section $t:U\to T$ has open image. Every map between two local homeomorphisms over $X$ is itself a local homeomorphism.

Proof. Around $t(x)$, choose a chart $W$ of $q$. Shrink to an open $V\subset U$ containing $x$ with $t(V)\subset W$. Then $t|_V=(q|_W)^{-1}|_V$, and $t(V)=(q|_W)^{-1}(V)$ is open in $T$. These images cover $t(U)$.

For a map $h:T\to T'$ over $X$, choose charts around $z$ and $h(z)$. Intersect the first with the inverse image of the second. Its projection is still open, and in these charts $h$ is the identity on that open subset of $X$. It is a homeomorphism onto an open subset of the second chart. Such neighborhoods cover $T$. $\square$

Theorem 2.2. If $T\to X$ is a local homeomorphism, evaluation is a natural homeomorphism over $X$: $\begin{gathered} \epsilon_T:E(\Gamma_X(T))\longrightarrow T,\\ [t]_x\longmapsto t(x). \end{gathered} \tag{2.1}$

Proof. Equal germs give equal values. Conversely, two sections with the same value at $x$ take values in a common chart after shrinking around $x$; there both are its inverse, so their germs agree. Every $z\in T$ is the value of the inverse of a chart at $q(z)$. Thus evaluation is bijective. On a basic set $[t,U]$, it is the section $t$ composed with its projection. This is continuous and, by Lemma 2.1, a homeomorphism onto the open image $t(U)$. Evaluation is therefore a homeomorphism. For $h:T\to T'$, both $\epsilon_{T'}E(\Gamma_X h)$ and $h\epsilon_T$ send $[t]_x$ to $h(t(x))$, proving naturality. $\square$

There is a natural presheaf map $\begin{gathered} \eta_P:P\longrightarrow\Gamma_X(E(P)),\\ s\longmapsto\widehat s. \end{gathered} \tag{2.2}$

Theorem 2.3. The map $\eta_P$ is an isomorphism exactly when $P$ is a sheaf. Consequently $E$ and $\Gamma_X$ restrict to mutually inverse equivalences between sheaves on $X$ and $\operatorname{Et}(X)$, with all their morphisms.

Proof. Suppose $P$ is a sheaf. Two sections with the same germs agree on a neighborhood of every point; separation then makes them equal. This proves injectivity on every open, including the empty open by its sheaf axiom. A continuous section $\sigma:U\to E(P)$ locally has image in a chart $[s,V]$. Since it is a section of $p_P$, on that smaller domain it is exactly $\widehat s$. On overlaps these local representatives have equal germs, so separation makes their restrictions equal. Gluing gives $s\in P(U)$ with $\sigma=\widehat s$. The empty domain uses its unique section. Thus $\eta_P$ is an isomorphism. Conversely, its target is a sheaf, so an isomorphic presheaf is a sheaf.

The natural isomorphisms (2.1) and (2.2) give the equivalence. Explicitly, a map of sheaves gives (1.3); a map of their germ spaces gives the map on sections, transported by (2.2). These operations are inverse because both act on a germ by the value of its represented local section. No bijectivity condition is imposed on a morphism. $\square$

3. The adjunction and the associated sheaf

Theorem 3.1. For every presheaf $P$ and every space $T\to X$, there is a natural bijection $\begin{gathered} \operatorname{Hom}_{\operatorname{Top}/X}(E(P),T)\\ \simeq\operatorname{Nat}(P,\Gamma_X(T)). \end{gathered} \tag{3.1}$

Proof. A map $k:E(P)\to T$ sends $s\in P(U)$ to the continuous section $k\widehat s$. This commutes with restrictions. Conversely, for $\alpha:P\to\Gamma_X(T)$, put $k_\alpha([s]_x)=\alpha_U(s)(x). \tag{3.2}$

If two representatives agree after restriction, naturality of $\alpha$ makes their sections agree there, so this is well defined. On $[s,U]$, the map is $\alpha_U(s)\circ p_P$, hence continuous. The charts are an open cover, so the whole map is continuous and lies over $X$. Applying the two constructions consecutively returns the original map or transformation at every germ or section. Their formulas commute with composition in both variables, proving naturality. $\square$

The unit is (2.2); the counit for an arbitrary $T$ is still evaluation (2.1), which need not be invertible. The two triangle identities can be checked without any abstract adjunction theorem. The composite $\epsilon_{E(P)}E(\eta_P)$ sends $[s]_x$ to $\widehat s(x)=[s]_x$. The composite $\Gamma_X(\epsilon_T)\eta_{\Gamma_X(T)}$ sends a section $t$ to the section $x\mapsto t(x)$. Both are identities.

Corollary 3.2. The associated sheaf of $P$ is $aP=\Gamma_X(E(P)). \tag{3.3}$

Proof. For a sheaf $F$, use its isomorphism $F\simeq\Gamma_X(E(F))$, then (3.1) and the equivalence of Theorem 2.3. These give a bijection from sheaf maps $aP\to F$ to presheaf maps $P\to F$. Its map is restriction along $\eta_P$, by the explicit constructions in (3.1). Therefore extensions exist and are unique, which is the sheafification universal property. The formulas respect maps of $P$. In particular this is canonically isomorphic, with its unit, to any other associated-sheaf construction. $\square$

Its stalks are exactly the original stalks. More precisely, evaluation and its inverse are $\begin{gathered} (aP)_x\longrightarrow P_x,\quad[\sigma]_x\longmapsto\sigma(x),\\\\ P_x\longrightarrow(aP)_x,\quad[s]_x\longmapsto[\widehat s]_x. \end{gathered} \tag{3.4}$

Evaluation is independent of a section representative. Around $x$, a section $\sigma$ lies in one chart and equals the corresponding $\widehat s$. This proves that the two maps are inverse, and that their definition commutes with presheaf maps. The second is the stalk map of the unit.

Corollary 3.3. A morphism of sheaves is an isomorphism exactly when all its stalk maps are bijective.

Proof. Under the equivalence it is a map of local homeomorphisms, hence an open map by Lemma 2.1. Stalkwise bijectivity says precisely that this map is a bijection on the underlying disjoint unions of fibres. A continuous open bijection is a homeomorphism: the inverse image under its inverse of an open set is its open image. The equivalence then gives an inverse sheaf map. The converse follows by taking germs of an inverse. $\square$

4. Pulling back the actual germ space

Let $f:X\to Y$ be continuous and $F$ a sheaf on $Y$. The projection below is a local homeomorphism: $f^*E(F)=X\times_Y E(F)\longrightarrow X. \tag{4.1}$

Indeed, the pullback of a chart over $U\subset Y$ is a chart over $f^{-1}U$. Let $L_fF=\Gamma_X(f^*E(F))$. For $G$ on $X$, put $(f_*G)(U)=G(f^{-1}U)$; the sheaf axiom follows by pulling back an open cover.

Theorem 4.1. The functor $L_f$ is left adjoint to $f_*$, and hence is the usual sheaf inverse image, with its actual stalk identification $(L_fF)_x\simeq F_{f(x)}. \tag{4.2}$

Proof. A sheaf map $L_fF\to G$ is, by Theorem 2.3, a map $k:f^*E(F)\to E(G)$ over $X$. For $s\in F(U)$, send $s$ to the section on $f^{-1}U$ whose germ at $x$ is $k(x,[s]_{f(x)})$. It is continuous, since it is the composite of $k$ with the pullback of $\widehat s$. It therefore corresponds to a unique element of $G(f^{-1}U)$. This gives a map $F\to f_*G$, natural under restrictions.

Conversely, for $b:F\to f_*G$, define $k_b(x,[s]_{f(x)})=[b_U(s)]_x. \tag{4.3}$

If representatives agree near $f(x)$, their images agree on the inverse image of that neighborhood, so the formula is well defined. On each pullback chart it is the section associated to $b_U(s)$, hence continuous. These charts cover. The two assignments are inverse by their germ formulas, and commute with maps of $F$ and $G$. This proves the adjunction. By (3.4) and Theorem 2.2, the stalk of the section sheaf of a local homeomorphism is its fibre. The fibre of (4.1) at $x$ is exactly $F_{f(x)}$, giving (4.2). $\square$

For comparison with the presheaf formula, define $H(V)=\mathop{\operatorname{colim}}_{\substack{U\text{ open in }Y\\\\f(V)\subset U}}F(U), \tag{4.4}$

with arrows given by restriction to smaller neighborhoods of $f(V)$. This index is nonempty and directed by intersection. For $W\subset V$, the same representatives define the restriction to $H(W)$. A representative $s\in F(U)$ gives the section $x\mapsto(x,[s]_{f(x)})$ of (4.1), hence a map $H\to L_fF$. Every section of (4.1) is locally of this form, since its second coordinate lies in a germ chart near each point. Two such representatives with the same value at $x$ agree on a smaller neighborhood of $f(x)$, and therefore agree in $H(W)$ after shrinking $W$ around $x$. These two observations prove surjectivity and injectivity of $H_x\to F_{f(x)}$, respectively. Formula (3.4) and Corollary 3.3 now identify $aH\simeq L_fF$, with this very comparison map. No openness assumption on $f$ was used.

Pulling back twice and pulling back along a composite identify the same triples of points. In charts their identification is a homeomorphism. Thus these are the usual coherent inverse-image comparisons, not just unrelated stalkwise bijections.

5. Finite limits and the limits that need étalification

Proposition 5.1. Finite limits of local homeomorphisms over $X$ are their ordinary limits in $\operatorname{Top}/X$. They correspond to finite limits of sheaves. Consequently stalks preserve finite limits.

Proof. The terminal object is $X\to X$. A fibre product is locally a common open subset of two charts, so is a local homeomorphism. A section of it is exactly a pair of sections with the prescribed equality. For parallel maps $h,k:T\rightrightarrows T'$ over $X$, their equality locus is open: near a point where they agree, put both images into the same chart of $T'$; equality of their projections then forces equality of the maps there. This open subspace of $T$ is the ordinary equalizer, and its sections are exactly the equalizer of the section maps. Finite products and equalizers construct every finite limit, by imposing all diagram equations on the finite product. These constructions establish the claimed correspondence. Fibres preserve these finite products and equalizers as sets, proving the stalk assertion. $\square$

For an arbitrary $T\to X$, call $E(\Gamma_X(T))\to T$ its étalification. Every map from a local homeomorphism $A\to X$ to $T$ factors uniquely through this evaluation map. In fact, apply (3.1) to the sheaf $\Gamma_X(A)$, use $E(\Gamma_X(A))\simeq A$, and then use the sheaf/étalé equivalence to obtain $\begin{gathered} \operatorname{Hom}_{\operatorname{Et}(X)}\bigl(A,E(\Gamma_X(T))\bigr)\\ \simeq\operatorname{Hom}_{\operatorname{Top}/X}(A,T). \end{gathered} \tag{5.1}$

The formulas in Section 3 show that the map in (5.1) is composition with evaluation. Thus étalification is right adjoint to the full inclusion $\operatorname{Et}(X)\subset\operatorname{Top}/X$.

An arbitrary small diagram in $\operatorname{Top}/X$ has a limit: take the subspace of the product consisting of tuples with a common base point and satisfying all diagram equations; for the empty diagram use $X$. Its usual projections have the universal property by the product and subspace topologies. For a diagram of local homeomorphisms, étalify this ordinary limit. Equation (5.1) says its maps from $A$ are exactly the compatible cones from $A$ to the original diagram. This is its limit in $\operatorname{Et}(X)$, and hence in sheaves. The ordinary infinite product may fail to be a local homeomorphism; étalification cannot in general be omitted.

6. Which étalé spaces are coverings?

For a set $S$, let $S_X(U)$ be the locally constant functions $U\to S$, where $S$ is discrete. Such functions glue uniquely and locally constantly, so $S_X$ is a sheaf. Evaluation identifies its stalk at $x$ with $S$: every representative is constant near $x$, and every value has a constant representative. Its germ charts are the local graphs of these functions. They give a natural homeomorphism $E(S_X)\simeq X\times S, \tag{6.1}$

because charts of constant sections are $U\times\{s\}$, which form the product basis, and locally constant graphs are their unions.

Say a sheaf is locally constant if it is isomorphic to $S_U$ on members of an open cover, with the set $S$ allowed to vary. A map is a covering here if every base point has a neighborhood $U$ over which it is $U\times S\to U$ for a discrete set $S$; empty fibres are allowed. If one's definition requires surjectivity, add that every stalk is nonempty.

Theorem 6.1. A sheaf is locally constant exactly when its germ space is a covering in this sense.

Proof. Restricting a germ space to $p^{-1}U$ is the germ space of the restricted sheaf, since representatives can be shrunk inside $U$; their chart topologies coincide. Formula (6.1) therefore sends each constant-sheaf trivialization to a covering trivialization. Conversely the sheaf of sections of $U\times S\to U$ is precisely $S_U$: sections are maps $x\mapsto(x,t(x))$, continuous exactly when the discrete-valued $t$ is locally constant. Apply the equivalence on every trivializing open. These arguments identify all maps as well as the objects. $\square$

The local-homeomorphism condition chooses a chart around one point of the total space. A covering chart must work for the entire fibre simultaneously. Neither constant infinite fibre cardinality nor the Hausdorff property supplies this stronger requirement, as Exercise 4 shows.

7. Four graded exercises with complete solutions

Exercise 1 (foundation: an explicit two-point germ space)

Let $X=\{0,1\}$ have opens $\varnothing,\{1\},X$. Describe all sheaves using a function $r:A\to B$. Give their germ spaces, every open subset of those spaces, and all their morphisms.

Solution. Put $F(X)=A$, $F(\{1\})=B$, $F(\varnothing)=\{*\}$, with restriction $r$ and the unique empty restrictions. Every cover of $X$ contains $X$, and every nonempty cover of $\{1\}$ contains $\{1\}$. Its compatible families therefore have exactly their specified section as glue. The empty cover has the unique empty section. Thus every $r$ gives a sheaf, and every sheaf gives exactly this data.

The smallest neighborhoods give $F_0=A$, $F_1=B$. Write the points as $a_0,b_1$. The basic sets are $\{a_0,r(a)_1\}$ and $\{b_1\}$. Hence an arbitrary subset, written $A'_0\amalg B'_1$, is open exactly when $r(A')\subset B'$: necessity follows from a basic neighborhood of each $a_0$, and sufficiency follows by taking the union of these basic sets and the remaining $b_1$ singletons. The projection of each two-point chart is a homeomorphism to $X$; each singleton chart maps to $\{1\}$.

A sheaf map is a pair $u:A\to A'$, $v:B\to B'$ with $vr=r'u$. It gives a map over $X$ by these two fibre maps. Conversely a continuous map over $X$ must send the basic neighborhood of $u(a)_0$ back to an open neighborhood of $a_0$, forcing $v(r(a))=r'(u(a))$. This proves the same commuting-square condition directly from topology.

Exercise 2 (intermediate: Hausdorff bases can have non-Hausdorff germ spaces)

For a sheaf $F$ on a Hausdorff space $X$, prove that $E(F)$ is Hausdorff exactly when every germ-equality locus of two sections on a common open is closed in that open. Apply the criterion to the sheaf of continuous real functions on $\mathbb R$.

Solution. If $E(F)$ is Hausdorff, its diagonal is closed: two unequal points have disjoint neighborhoods, whose product misses the diagonal. For sections $s,t\in F(U)$, the continuous map $x\mapsto(\widehat s(x),\widehat t(x))$ pulls this diagonal back to their germ-equality locus. Thus it is closed in $U$.

Conversely, points over distinct base points have disjoint inverse images of base neighborhoods. For unequal germs over the same $x$, choose section representatives on a common open $U$. The closed equality locus misses $x$, so on an open neighborhood $V\subset U$ of $x$ it is empty. The two chart images over $V$ are disjoint, since their intersection would be an equality at the same base point. They separate the two germs. This proves the criterion.

Continuous functions form a sheaf by gluing functions and checking continuity on the open cover. Take $s(x)=0$ and $t(x)=\max(x,0)$. Their germs agree exactly for $x<0$: at $0$ every neighborhood contains positive points where the functions differ. The equality locus $(-\infty,0)$ is not closed in $\mathbb R$. The two germs at $0$ cannot be separated, since any pair of their chart neighborhoods intersects over negative points sufficiently close to $0$. Thus the germ space is not Hausdorff.

Let $X=\{0\}\cup\{1/n:n\ge1\}\subset\mathbb R$, and let $F_n=\{0,1\}_X$ for every $n\ge1$. Show that the natural map from the stalk of the product sheaf to the product of the stalks at $0$ is surjective but not injective. Explain why the ordinary product of their germ spaces is not étalé.

Solution. Products of sheaves are computed on opens: compatible tuples glue coordinate by coordinate, uniquely. Each singleton $\{1/n\}$ is both open and closed in $X$. Its characteristic function $s_n:X\to\{0,1\}$ is therefore locally constant. Every $s_n$ is zero on a neighborhood of $0$, but there is no common neighborhood of $0$ on which all are zero. Hence the germ of the tuple $(s_n)_n$ differs from the zero tuple's germ, while each of its coordinate germs equals zero. This proves noninjectivity. Every tuple of stalk values is represented by its tuple of globally constant functions, proving surjectivity.

By (6.1), the ordinary fibre product of all the germ spaces is $X\times\{0,1\}^{\mathbb N}$, with the product topology on the second factor. Its fibre over $0$ has no isolated point: a basic product neighborhood fixes only finitely many coordinates, and one may change another. A fibre of a local homeomorphism is discrete, since each chart meets it in just one point. Thus this product is not a local homeomorphism. Its étalification is the germ space of the product sheaf, and evaluation has exactly the failure of fibrewise injectivity just exhibited. Finite-limit preservation from Proposition 5.1 gives no assertion for this product.

Exercise 4 (synthesis: constant cardinality without a covering)

Let $E$ be the topological disjoint union of countably many copies of $\mathbb R$ and one copy of $(0,\infty)$, with each component projected to $\mathbb R$ by its usual inclusion. Prove that the projection is a surjective local homeomorphism between Hausdorff spaces, has countably infinite fibres everywhere, and is not a covering. Deduce a sheaf with constant stalk cardinality which is not locally constant.

Solution. Each component is open and maps homeomorphically onto an open subset of $\mathbb R$, so the projection is a local homeomorphism. The full copies make it surjective. Distinct components are disjoint open sets; within one component the usual Hausdorff neighborhoods separate points. Thus the disjoint union is Hausdorff. Every nonpositive point has one preimage in each full copy, while a positive point has one additional preimage. Both fibres are countably infinite.

Suppose an evenly covered neighborhood of $0$ existed. Restrict it to a connected open interval $V$ around $0$. Each sheet over $V$ is connected because it is homeomorphic to $V$. A connected subset of this disjoint union lies in a single component: meeting one component and its open complement would disconnect it. Every sheet has a point over $0$, so it must lie in a full copy of $\mathbb R$, not in the extra positive component. Consequently their union cannot cover the points of that component lying over positive points of $V$. This contradicts the covering property.

Take $F=\Gamma_{\mathbb R}(E)$. Its stalks are the fibres by Theorem 2.2 and (3.4), so all have the same countably infinite cardinality. Its germ space is $E$. Theorem 6.1 and the proved failure of a covering imply that $F$ is not locally constant. No finiteness hypothesis can be silently replaced by a cardinality assertion here.