02.01.E1 · analysis / topology

Point-set topology and the fundamental groupoid exercise pack (Brown, Topology and Groupoids supplement)

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Formal definition of the pack Intermediate

Ronald Brown's Topology and Groupoids develops algebraic topology from the point-set foundations through to the fundamental groupoid $π_{1} (X, A)$ on a chosen set $A \subseteq X$ of base points, rather than the fundamental group $π_{1} (X, x_{0})$ at a single point. Brown's distinctive editorial choices are: the identification (quotient) topology presented universal-property-first; the Seifert-van Kampen theorem proved in its groupoid form, which needs no connectivity hypothesis on the intersection $U \cap V$ ; and covering spaces treated as a representation theory of the fundamental groupoid, making the Galois correspondence explicit.

This pack collects eleven exercises drawn from Brown's chapters — three easy, five medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The problems are grouped by Brown's progression: the point-set layer (open and closed sets, continuity, compactness, connectedness), the identification-topology layer (quotients, cones, adjunction spaces), the fundamental group(oid) and van Kampen layer, and the covering-space layer.

The notation throughout follows the Babel Bible convention recorded in the fundamental-groupoid unit: $π_{1} (X, x_{0})$ denotes the fundamental group at the base point $x_{0}$ , and $π_{1} (X, A)$ denotes the fundamental groupoid on the set $A \subseteq X$ , whose objects are the points of $A$ and whose morphisms are homotopy classes of paths rel endpoints. Brown himself writes $π X$ for the groupoid on the whole space; some authors write $Π_{1}$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining ten follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Compute $π_{1} (S^{1})$ using the groupoid form of the Seifert-van Kampen theorem.

Solution. This is the computation that motivates the groupoid form. Cover the circle by two open arcs $U$ and $V$ , each an arc slightly larger than a semicircle, chosen so that $U \cup V = S^{1}$ and $U \cap V$ is the disjoint union of two open arcs, $W_{1}$ and $W_{2}$ (the two regions where the arcs overlap). Each of $U$ , $V$ , $W_{1}$ , $W_{2}$ is contractible.

The group form of Seifert-van Kampen does not apply directly: it requires $U \cap V$ to be path-connected, and here $U \cap V = W_{1} ⊔ W_{2}$ is not. The groupoid form has no such requirement. Choose a set of base points $A = {a, b}$ with $a \in W_{1}$ and $b \in W_{2}$ , one in each component of the intersection. The theorem states that the fundamental groupoid $π_{1} (S^{1}, A)$ is the pushout of groupoids

π_{1} (S^{1}, A) = π_{1} (U, A) *_{π_{1} (U \cap V, A)} π_{1} (V, A) .

Compute each piece on the base-point set $A = {a, b}$ :

$π_{1} (U, A)$ is the groupoid on two points in a contractible space: a single isomorphism class, with a unique morphism $a \to b$ . As a groupoid it is the "indiscrete" groupoid on two objects, often written $I$ . Likewise $π_{1} (V, A) ≅ I$ .
$π_{1} (U \cap V, A)$ : here $a, b$ lie in different components $W_{1}, W_{2}$ , so there are no paths between them. The groupoid is discrete on two objects — only the two identity morphisms.

The pushout glues two copies of $I$ (each supplying an arrow $a \to b$ ) along the discrete groupoid on ${a, b}$ . Call the two glued arrows $u : a \to b$ (through $U$ ) and $v : a \to b$ (through $V$ ). The composite $v^{- 1} u : a \to a$ is a non-identity loop, and the resulting groupoid is freely generated by $u$ and $v$ subject to no relations between the two generating arrows. The vertex group at $a$ is therefore the free group on the single generator $g = v^{- 1} u$ , namely $Z$ .

Restricting the groupoid $π_{1} (S^{1}, A)$ to the single object $a$ recovers the fundamental group:

π_{1} (S^{1}, a) = Z . □

The generator $g$ winds once around the circle; its $n$ -th power winds $n$ times. The point of the groupoid form is that it handles the disconnected intersection $W_{1} ⊔ W_{2}$ with no special pleading, where the group form must first be repaired by an ad hoc argument. This is Brown's headline application.

Exercises Intermediate

Exercise 1 (easy, proof). Continuity via closed sets.

Let $f : X \to Y$ be a map between topological spaces. Show that $f$ is continuous if and only if $f^{- 1} (C)$ is closed in $X$ for every closed set $C \subseteq Y$ .

Hint

Preimage commutes with complement: $f^{- 1} (Y ∖ C) = X ∖ f^{- 1} (C)$ . Continuity says preimages of opens are open.

Answer

Suppose $f$ is continuous and $C \subseteq Y$ is closed. Then $Y ∖ C$ is open, so $f^{- 1} (Y ∖ C)$ is open in $X$ . Preimage commutes with complementation:

f^{- 1} (Y ∖ C) = X ∖ f^{- 1} (C),

so $X ∖ f^{- 1} (C)$ is open, hence $f^{- 1} (C)$ is closed.

Conversely, suppose $f^{- 1} (C)$ is closed for every closed $C$ . Let $O \subseteq Y$ be open; then $Y ∖ O$ is closed, so $f^{- 1} (Y ∖ O) = X ∖ f^{- 1} (O)$ is closed, hence $f^{- 1} (O)$ is open. So $f$ is continuous. The two formulations are equivalent because the open sets and the closed sets each determine the other by complementation. Brown §3.

Exercise 2 (easy, proof). Continuous image of a connected space.

Let $f : X \to Y$ be continuous and surjective. Show that if $X$ is connected, then $Y$ is connected.

Hint

A space is connected iff its only clopen subsets are $\emptyset$ and the whole space. Pull back a clopen set of $Y$ .

Answer

Suppose for contradiction that $Y$ is disconnected: $Y = A ⊔ B$ with $A, B$ non-empty, open, and disjoint. Then $f^{- 1} (A)$ and $f^{- 1} (B)$ are open (continuity), disjoint, and cover $X$ since $A \cup B = Y$ . Both are non-empty because $f$ is surjective. So $X = f^{- 1} (A) ⊔ f^{- 1} (B)$ exhibits a disconnection of $X$ — contradicting connectedness.

Therefore $Y$ is connected. Equivalently: connectedness is a topological invariant preserved by continuous surjections, so it descends along quotient maps. This is the engine behind the intermediate value theorem (the image of an interval under a continuous real map is an interval). Brown §3.6.

Exercise 4 (medium, proof). The quotient by the universal property.

Let $q : X \to Y$ be a quotient (identification) map, and let $g : X \to Z$ be a continuous map that is constant on the fibres of $q$ (i.e.\ $q (x) = q (x^{'})$ implies $g (x) = g (x^{'})$ ). Show that there is a unique continuous map $\overset{g}{ˉ} : Y \to Z$ with $\overset{g}{ˉ} \circ q = g$ .

Hint

Define $\overset{g}{ˉ}$ set-theoretically by $\overset{g}{ˉ} (y) = g (x)$ for any $x \in q^{- 1} (y)$ . Use the defining property of the quotient topology — a set in $Y$ is open iff its preimage under $q$ is open — to get continuity.

Answer

Existence and uniqueness as a function. For $y \in Y$ , pick any $x$ with $q (x) = y$ (one exists since $q$ is surjective as a quotient map) and set $\overset{g}{ˉ} (y) := g (x)$ . This is well defined: if $q (x) = q (x^{'}) = y$ , then $g (x) = g (x^{'})$ by hypothesis. Any map $\overset{g}{ˉ}$ satisfying $\overset{g}{ˉ} \circ q = g$ must take this value, so $\overset{g}{ˉ}$ is unique.

Continuity. Let $O \subseteq Z$ be open. Then

q^{- 1} (\overset{g}{ˉ}^{- 1} (O)) = (\overset{g}{ˉ} \circ q)^{- 1} (O) = g^{- 1} (O),

which is open in $X$ because $g$ is continuous. By the defining property of the quotient topology on $Y$ , a subset $S \subseteq Y$ is open precisely when $q^{- 1} (S)$ is open. Taking $S = \overset{g}{ˉ}^{- 1} (O)$ , we conclude $\overset{g}{ˉ}^{- 1} (O)$ is open. So $\overset{g}{ˉ}$ is continuous.

This universal property is exactly how Brown organises identification topology: the quotient is characterised as the initial object through which fibre-constant maps factor. It is the tool that gives well-defined continuous maps out of cones, mapping cones, and adjunction spaces. Brown §4.

Exercise 5 (medium, proof). The cone on any space is contractible.

Let $X$ be a topological space. The cone is $C X = (X \times [0, 1]) / (X \times {1})$ , collapsing the top to a single point $v$ . Show that $C X$ is contractible.

Hint

Slide every point along its line toward the cone vertex. Write the homotopy on $X \times [0, 1]$ first, then check it descends to the quotient.

Answer

Write points of $C X$ as $[x, t]$ with $[x, 1] = v$ for all $x$ . Define $H : C X \times [0, 1] \to C X$ by

H ([x, t], s) = [x, t + s (1 - t)] .

At $s = 0$ , $H ([x, t], 0) = [x, t]$ , the identity. At $s = 1$ , $H ([x, t], 1) = [x, 1] = v$ , the constant map to the vertex. So $H$ is a homotopy from $id_{C X}$ to the constant map at $v$ .

Continuity. The map $\tilde{H} : (X \times [0, 1]) \times [0, 1] \to C X$ , $\tilde{H} ((x, t), s) = [x, t + s (1 - t)]$ , is continuous (composition of continuous operations with the quotient map). It is constant on the fibres of the quotient $X \times [0, 1] \to C X$ at every fixed $s$ : when $t = 1$ the value is $v$ regardless of $x$ . By the universal property of the quotient topology (Exercise 4, applied with the parameter $s$ ), $\tilde{H}$ descends to a continuous $H$ on $C X \times [0, 1]$ .

Hence $C X$ deformation-retracts to $v$ and is contractible. In particular every space embeds in a contractible space (as the base $X \times {0} \subset C X$ ), which is why cones are the building block for mapping cones and suspensions. Brown §4.

Exercise 6 (medium, symbolic). Fundamental group of the wedge of two circles.

Use the Seifert-van Kampen theorem to compute $π_{1} (S^{1} \lor S^{1})$ , the figure-eight.

Hint

Take $U$ and $V$ to be open neighbourhoods of the two loops, each deformation-retracting to one circle, with $U \cap V$ contractible (a neighbourhood of the wedge point). Then van Kampen gives a free product.

Answer

Let $X = S^{1} \lor S^{1}$ with wedge point $x_{0}$ . Choose $U$ = the first circle together with a small open arc of the second around $x_{0}$ , and $V$ = the second circle together with a small open arc of the first around $x_{0}$ . Then $U$ deformation-retracts to the first circle, so $π_{1} (U, x_{0}) = Z = ⟨ a ⟩$ ; likewise $π_{1} (V, x_{0}) = Z = ⟨ b ⟩$ . The intersection $U \cap V$ is a small open neighbourhood of $x_{0}$ shaped like a plus sign, which is contractible, so $π_{1} (U \cap V, x_{0}) = 1$ .

The Seifert-van Kampen theorem gives the amalgamated free product over the one-element group:

π_{1} (S^{1} \lor S^{1}, x_{0}) = Z *_{1} Z = Z * Z = F_{2},

the free group on two generators $a, b$ . The generators are the two loops; there are no relations because the intersection is simply connected. More generally $π_{1} (⋁_{i = 1}^{n} S^{1}) = F_{n}$ , the free group on $n$ generators. This is the basic non-abelian fundamental group and the reason graphs have free fundamental groups. Brown §6.7; Hatcher §1.2.

Exercise 7 (medium, symbolic). Fundamental group of the torus.

Compute $π_{1} (T^{2})$ where $T^{2} = S^{1} \times S^{1}$ , and identify it as an explicit group.

Hint

Either use the product rule $π_{1} (X \times Y) = π_{1} (X) \times π_{1} (Y)$ , or present the torus as a square with edges identified and run van Kampen with $U$ the interior and $V$ a neighbourhood of the edge graph.

Answer

Via the product rule. The fundamental group of a product is the product of fundamental groups (paths in $X \times Y$ are pairs of paths). So

π_{1} (T^{2}) = π_{1} (S^{1}) \times π_{1} (S^{1}) = Z \times Z = Z^{2} .

Via van Kampen on the square model. Present $T^{2}$ as the square $[0, 1]^{2}$ with $ab a^{- 1} b^{- 1}$ edge identifications, where $a, b$ are the two generating loops. Let $V$ be a neighbourhood of the edge graph (a wedge of two circles, $π_{1} = F_{2} = ⟨ a, b ⟩$ ) and $U$ the open disc interior (contractible). The intersection $U \cap V$ is an annulus, $π_{1} = Z$ , generated by a loop that maps to the boundary word $ab a^{- 1} b^{- 1}$ in $π_{1} (V)$ and to $1$ in $π_{1} (U)$ . Van Kampen yields

π_{1} (T^{2}) = ⟨ a, b ∣ ab a^{- 1} b^{- 1} = 1 ⟩ = ⟨ a, b ∣ ab = ba ⟩ ≅ Z^{2} .

Both routes agree: the torus has abelian fundamental group $Z^{2}$ , the single relation $ab = ba$ recording that the two loops commute. Brown §6.7; Hatcher §1.2.

Exercise 8 (medium, proof). Why the groupoid van Kampen needs no connectivity.

Explain, with the circle as the worked case, why the group form of the Seifert-van Kampen theorem fails to apply when $U \cap V$ is disconnected, and how the groupoid form repairs this.

Hint

The group form is a pushout of groups at one base point; it needs a single base point in a path-connected $U \cap V$ so loops can be transported. With a disconnected intersection, choose one base point per component and work in the groupoid.

Answer

The group form of Seifert-van Kampen states: if $X = U \cup V$ with $U, V, U \cap V$ open and path-connected and containing a common base point $x_{0}$ , then $π_{1} (X, x_{0})$ is the pushout (amalgamated free product) $π_{1} (U) *_{π_{1} (U \cap V)} π_{1} (V)$ . The path-connectedness of $U \cap V$ is essential: the relations come from loops in $U \cap V$ based at $x_{0}$ , and if $U \cap V$ is disconnected, a single base point cannot see the other components, so loops crossing between components are invisible to the group pushout.

For the circle, the only natural two-set cover by contractible arcs has $U \cap V$ equal to two disjoint arcs $W_{1} ⊔ W_{2}$ . The single-base-point group pushout would compute $π_{1} (U) *_{π_{1} (U \cap V)} π_{1} (V) = 1 *_{1} 1 = 1$ , falsely giving the one-element group — it misses the generating loop entirely, because that loop is built by going out through $W_{1}$ and back through $W_{2}$ , a path that the single-component group form cannot assemble.

The groupoid form fixes this by allowing a set of base points: choose $a \in W_{1}$ , $b \in W_{2}$ . The fundamental groupoid $π_{1} (X, {a, b})$ is the pushout of groupoids, and the morphism from $a$ to $b$ through $U$ , composed with the inverse of the morphism through $V$ , is the generating loop $g \in π_{1} (S^{1}, a) = Z$ . The groupoid records the inter-component paths that the group form discards. This is the precise sense in which Brown's reformulation is "the right one": it removes a hypothesis that the geometry routinely violates. Brown §6.7; Brown (1967).

Exercise 9 (hard, proof). Lifting criterion for covering spaces.

Let $p : \tilde{X} \to X$ be a covering map with $\tilde{X}$ path-connected, and let $f : Y \to X$ be continuous with $Y$ path-connected and locally path-connected. Fix base points with $f (y_{0}) = x_{0}$ and $p (\tilde{x}_{0}) = x_{0}$ . Show that a lift $\tilde{f} : Y \to \tilde{X}$ with $\tilde{f} (y_{0}) = \tilde{x}_{0}$ exists if and only if $f_{*} π_{1} (Y, y_{0}) \subseteq p_{*} π_{1} (\tilde{X}, \tilde{x}_{0})$ .

Hint

For necessity, push the inclusion $\tilde{f} \circ$ through functoriality. For sufficiency, define $\tilde{f} (y)$ by lifting the image path $f \circ γ$ from $y_{0}$ to $y$ and checking the endpoint is independent of $γ$ — this is where the subgroup condition is used, via unique path lifting.

Answer

Necessity. If a lift $\tilde{f}$ exists with $p \circ \tilde{f} = f$ , then applying $π_{1}$ and functoriality, $f_{*} = p_{*} \circ \tilde{f}_{*}$ , so

f_{*} π_{1} (Y, y_{0}) = p_{*} (\tilde{f}_{*} π_{1} (Y, y_{0})) \subseteq p_{*} π_{1} (\tilde{X}, \tilde{x}_{0}) .

Sufficiency. Assume $f_{*} π_{1} (Y, y_{0}) \subseteq p_{*} π_{1} (\tilde{X}, \tilde{x}_{0})$ . For $y \in Y$ , choose a path $γ$ from $y_{0}$ to $y$ (it exists since $Y$ is path-connected). Then $f \circ γ$ is a path in $X$ from $x_{0}$ ; by the path-lifting property of covering maps it lifts uniquely to a path $f \circ γ$ in $\tilde{X}$ starting at $\tilde{x}_{0}$ . Define $\tilde{f} (y) := f \circ γ (1)$ , the endpoint of the lift.

Well-definedness. Suppose $γ^{'}$ is another path from $y_{0}$ to $y$ . Then $γ \cdot γ^{' - 1}$ is a loop at $y_{0}$ , so $[f \circ (γ \cdot γ^{' - 1})] = f_{*} [γ \cdot γ^{' - 1}] \in f_{*} π_{1} (Y, y_{0}) \subseteq p_{*} π_{1} (\tilde{X}, \tilde{x}_{0})$ . A class lies in $p_{*} π_{1} (\tilde{X}, \tilde{x}_{0})$ exactly when its lift starting at $\tilde{x}_{0}$ is a loop (closes up). Hence the lift of $f \circ (γ \cdot γ^{' - 1})$ is a loop, which forces the lifts of $f \circ γ$ and $f \circ γ^{'}$ to share the same endpoint. So $\tilde{f} (y)$ does not depend on the chosen path.

Continuity. Local path-connectedness of $Y$ and the local-homeomorphism property of $p$ make $\tilde{f}$ continuous: near any $y$ , choose an evenly covered neighbourhood of $f (y)$ , the local inverse of $p$ composed with $f$ agrees with $\tilde{f}$ , and these local descriptions are continuous.

Thus the lift exists iff the subgroup condition holds. This lifting criterion is the algebraic heart of covering-space theory; specialised to $Y = \tilde{X}^{'}$ another covering, it yields the classification of coverings by conjugacy classes of subgroups of $π_{1} (X, x_{0})$ . Brown §10; Hatcher §1.3.

Exercise 10 (hard, symbolic). Classify the connected coverings of the figure-eight up to two sheets.

Using the Galois correspondence, classify the connected covering spaces of $S^{1} \lor S^{1}$ with at most two sheets, and for each give the corresponding subgroup of $π_{1} = F_{2} = ⟨ a, b ⟩$ .

Hint

$n$ -sheeted connected coverings correspond to index- $n$ subgroups of $π_{1}$ , up to conjugacy. The number of sheets equals the index. Subgroups of index $\leq 2$ in $F_{2} = ⟨ a, b ⟩$ : the whole group (index 1) and the kernels of surjections $F_{2} \to Z /2$ .

Answer

By the Galois correspondence, connected $n$ -sheeted coverings of $X = S^{1} \lor S^{1}$ correspond to index- $n$ subgroups of $π_{1} (X) = F_{2} = ⟨ a, b ⟩$ , with conjugate subgroups giving isomorphic coverings (and normal subgroups giving regular/normal coverings).

One sheet (index 1). The whole group $F_{2}$ itself. The covering is the identity $X \to X$ .

Two sheets (index 2). Index-2 subgroups are automatically normal, and they are exactly the kernels of the surjective homomorphisms $F_{2} \to Z /2$ . Such a homomorphism sends each generator to $0$ or $1$ in $Z /2$ , and must be surjective (not both $0$ ). The three surjections are:

$a \mapsto 1, b \mapsto 0$ : kernel $H_{1} = ⟨ a^{2}, b, ab a^{- 1} ⟩$ . The covering is the graph where the $a$ -loop is doubled into a length-two cycle and each vertex carries a $b$ -loop.
$a \mapsto 0, b \mapsto 1$ : kernel $H_{2}$ , the symmetric case with roles of $a$ and $b$ swapped.
$a \mapsto 1, b \mapsto 1$ : kernel $H_{3} = ⟨ a^{2}, b^{2}, ab ⟩$ . The covering is the graph where both loops are doubled and interleaved into a single 4-edge cycle alternating $a, b$ .

So there are three distinct connected double covers, one for each index-2 subgroup, all normal (deck group $Z /2$ ). Each total space is a connected 4-valent graph on two vertices, and since coverings of graphs are graphs, each has free fundamental group — here of rank $1 + 2 (2 - 1) = 3$ by the Euler-characteristic count $χ (\tilde{X}) = n \cdot χ (X)$ , $χ (X) = 1 - 2 = - 1$ , so $χ (\tilde{X}) = - 2$ and rank $= 1 - χ = 3$ , matching the three listed generators of each kernel. Brown §9-§10; Hatcher §1.3.

Exercise 11 (hard, proof). A continuous bijection that is not a homeomorphism, and when compactness saves it.

(a) Give a continuous bijection $f : X \to Y$ that is not a homeomorphism. (b) Prove that if $X$ is compact and $Y$ is Hausdorff, then every continuous bijection $f : X \to Y$ is a homeomorphism.

Hint

For (a), use the half-open interval $[0, 2 π)$ wrapping onto the circle. For (b), show $f$ is a closed map, using "closed in compact is compact" and "compact in Hausdorff is closed", then a closed continuous bijection has continuous inverse.

Answer

(a) Let $X = [0, 2 π)$ with the subspace topology from $R$ , $Y = S^{1}$ , and $f (t) = (cos t, sin t)$ . This is a continuous bijection. It is not a homeomorphism: the inverse is discontinuous at the point $(1, 0)$ , where points just below $2 π$ map near $0$ but the inverse sends a neighbourhood of $(1, 0)$ to a set that is not open in $[0, 2 π)$ (it splits across $0$ and the missing endpoint). Concretely, $f ([0, π))$ is an open arc's worth of $X$ but its image is not open in $S^{1}$ , so $f$ is not an open map. Here $X$ is not compact, which is what allows the failure.

(b) Suppose $X$ is compact and $Y$ Hausdorff, and $f : X \to Y$ is a continuous bijection. To show $f^{- 1}$ is continuous, it suffices to show $f$ is a closed map (then preimages under $f^{- 1}$ , which are images under $f$ , of closed sets are closed). Let $C \subseteq X$ be closed. Then:

$C$ is closed in the compact space $X$ , hence compact (Exercise 3).
The continuous image $f (C)$ is compact (continuous images of compact sets are compact).
A compact subset of the Hausdorff space $Y$ is closed.

So $f (C)$ is closed for every closed $C$ ; $f$ is a closed map. A closed continuous bijection has continuous inverse: for the inverse $g = f^{- 1}$ and any closed $C \subseteq X$ , $g^{- 1} (C) = f (C)$ is closed, so $g$ is continuous. Therefore $f$ is a homeomorphism.

The contrast with (a) is the lesson: continuity of the inverse is automatic exactly when the domain's compactness and the codomain's Hausdorffness conspire to make $f$ closed. Brown §3.5-§3.6; Munkres §26.

Exercise pack supplementing Ronald Brown, Topology and Groupoids: point-set topology (§3), identification spaces (§4), the fundamental group and groupoid and the Seifert-van Kampen theorem in groupoid form (§6), and covering spaces as fundamental-groupoid representations (§9-§10). Eleven exercises: 3 easy, 5 medium, 3 hard.

Prerequisites

02.01.01 pending
02.01.06 pending
03.12.00 pending
03.12.08 pending
03.12.09 pending
03.12.02 pending

Tier anchors

beginner
intermediate: Brown, Topology and Groupoids (3rd ed.), exercises across §3 (open/closed sets, continuity), §3.5-§3.6 (compactness, connectedness), §4 (identification spaces), §6 (fundamental group and groupoid), §6.7 (Seifert-van Kampen, groupoid form), §10 (covering spaces)
master

References

Brown — Topology and Groupoids (3rd ed., BookSurge 2006) · Exercises in §3 (point-set: open/closed, continuity, compactness, connectedness), §4 (identification/quotient topology: cones, mapping cones, adjunction spaces), §6 (fundamental group $\pi_1(X,x_0)$ and fundamental groupoid $\pi_1(X,A)$), §6.7 (Seifert-van Kampen theorem, groupoid form), §9-§10 (covering spaces as groupoid representations)
Brown — Groupoids and van Kampen's theorem, Proc. London Math. Soc. 17 (1967) 385-401 · Original groupoid Seifert-van Kampen theorem
Hatcher — Algebraic Topology · §1.1-§1.3 (fundamental group, van Kampen, covering spaces); §0 (quotient constructions)
Munkres — Topology (2nd ed.) · Ch. 3-4 (connectedness, compactness, separation); Ch. 9 (fundamental group, covering spaces)

Estimated time

intermediate: 5h