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We use planar duality for rectangle crossings, Harris-FKG inequality and the permitted exponential decay of subcritical percolation. A critical-point detail matters: uniqueness stated only for $p>p_c$ cannot by itself be applied at $p=p_c$. We prove the Burton-Keane theorem by the boundary counting proof of percolation uniqueness, establishing at most one infinite percolation cluster at every parameter and avoiding that gap.

For $0<p<1$, the number $N$ of infinite percolation clusters is almost surely constant by translation ergodicity of Bernoulli percolation. It cannot be a finite constant greater than one. A sufficiently large box meets two different infinite percolation clusters with positive probability; opening its finitely many interior edges joins them without creating a new infinite percolation cluster, decreasing $N$. Finite modification of Bernoulli percolation gives positive probability to this modification, contradicting constancy.

Nor can $N=\infty$. A finite box then meets three different infinite percolation clusters with positive probability. Select one infinite exterior branch from each. In the box and a finite collar retain just an embedded three-armed tree joining those branches and close all other incident edges there. Its branching graph vertex separates three infinite components when removed, so is a trifurcation vertex in percolation. All selections concern finitely many possible entrances and edge patterns; finite-energy property of Bernoulli percolation gives positive probability for at least one such pattern. Translation invariance then gives a common positive trifurcation probability $q$.

Here is the counting contradiction in detail. For a finite box $W$, contract each open component outside $W$ that touches $W$ to a boundary terminal. In each cluster, the resulting incidence graph is finite and connected. Every trifurcation in $W$ separates at least three sets of terminals, since each infinite branch must leave $W$. A minimal subtree joining all terminals must therefore contain that graph vertex with degree of a vertex at least three. Its leaves are terminals. The tree identity ${\sum }_v(\mathrm{deg}(v)-2)=-2$ bounds the number of those branching graph vertices by the terminal count. Different terminals are represented by different exterior graph neighbours of $W$. Thus the trifurcation boundary-counting lemma givesFor $W=[-r,r{]}^2\cap {\mathbb{Z}}^2$, the two sizes are $(2r+1{)}^2$ and $8r+4$. Letting $r\to \infty$ contradicts $q>0$. Therefore $N\in \{0,1\}$ almost surely, with the same conclusion for the translated planar dual graph.

Now prove absence of percolation at $p=1/2$ by Zhang's argument. Suppose instead that $\theta (1/2)>0$. Almost surely an infinite primal cluster exists, and self-duality gives an infinite dual cluster as well. An infinite connected subgraph of this locally finite lattice contains a graph ray, by the König infinity lemma. Expanding square boxes meet the infinite percolation clusters with probability tending to one.

For $B_r=[-r,r{]}^2\cap {\mathbb{Z}}^2$, let $A_N,A_E,A_S,A_W$ mean that an open graph ray takes an outward edge through the indicated side and subsequently uses graph vertices outside $B_r$. The union occurs whenever the box meets an infinite percolation cluster: take the last exit of an infinite graph ray from the finite box. Conversely an exterior arm supplies an infinite percolation cluster meeting its boundary. Quarter-turn symmetry makes the four probabilities equal. Their complements are decreasing events, so the square-root trick for positively associated events givesFor the planar dual graph use the box with boundary coordinates $\pm (r+1/2)$ and define its exterior side-arm events in the same way. This box also has quarter-turn symmetry and meets the dual infinite percolation cluster with probability tending to one. Thus each dual side has an infinite exterior arm with probability tending to one. Primal outward edges cross this dual-box contour on the corresponding sides; their remaining graph vertices stay outside it. A union bound makes the simultaneous event of primal north/south arms and dual east/west arms have positive probability for a sufficiently large box.

The four arms alternate around the contour. Open all edges with both endpoints in $B_r$, leaving all outward and exterior edges intact. This joins the two primal entrance points and preserves all four exterior arms. Finite-energy property of Bernoulli percolation keeps the event's probability positive. The primal north and south arms are now connected through the box. The two dual arms must lie in different infinite dual clusters. To see the separation, a hypothetical finite dual graph path joining the east and west arms cannot enter the dual-box interior: its boundary-crossing and interior edges cross primal edges that were just opened. Erase its loops and cut it at successive hits of the contour. A resulting exterior dual crosscut, together with the corresponding contour arc, encloses one of the two intervening primal entrance points. The infinite primal graph ray from that point would have to cross a dual-open edge, which is impossible. This is the planar separation used in the alternating arms argument at the self-dual percolation parameter. It contradicts uniqueness of the infinite dual cluster. HenceFinally suppose $p_c(2)>1/2$. At $p=1/2$ the permitted exponential decay of subcritical percolation would supply $C,c>0$ with ${\mathbb{P}}_{1/2}(v\leftrightarrow \partial B_n(v))\le C{e}^{-cn}$. Use the rectangle with graph vertices $\{0,\dots ,n\}\times \{0,\dots ,n-1\}$, and let $H_n$ be its left-to-right open crossing. Planar duality for rectangle crossings identifies its complement with a dual top-to-bottom crossing of a rectangle of width $n-1$ and height $n$. Rotation and translation give the same crossing law; side edges at the entrance and exit boundaries are irrelevant. Thus the exact self-dual rectangle crossing probability isBut a crossing has some starting graph vertex on its $n$-vertex left side connected to distance $n$. The union bound and exponential decay of subcritical percolation imply ${\mathbb{P}}_{1/2}(H_n)\le nC{e}^{-cn}\to 0$, a contradiction. ThereforeNo Russo-Seymour-Welsh theorem or continuity assertion for $\theta$ is being used, and critical uniqueness was proved rather than inferred from the supercritical hypothesis.

Fix an edge $e=\{u,v\}$ and orient it from $u$ to $v$. For each spanning tree $T$, let ${\gamma }_T$ be its unique graph path from $u$ to $v$. Assign a signed unit graph path flow $j_T$ to oriented edges: value $+1$ in the direction used by ${\gamma }_T$, value $-1$ on the reversed orientation, and zero otherwise. Define its meanEach $j_T$ has net divergence one at $u$, minus one at $v$ and zero elsewhere, so $J$ is a unit flow. The mean spanning-tree path current also satisfies the Kirchhoff cycle law. Although its verification is permitted to be omitted, a short counting argument makes it explicit. Let ${\mathcal{F}}_{uv}$ be the two-component spanning forests separating $u$ and $v$, write $A_F$ for the component containing $u$, and let $N_T$ be the number of spanning trees. Deleting an edge used by the oriented tree graph path gives one such forest. Conversely adjoining an edge across its two components gives a unique tree with that edge on its $u$-to-$v$ graph path. ConsequentlyThe gradient form proves the Kirchhoff cycle law directly and is Ohm's law for unit edge resistance. The Kirchhoff node law then gives $Lh={\delta }_u-{\delta }_v$, and the electrical unit flow is unique: the difference between two such voltages is a harmonic function on a graph and hence constant on the connected finite graph.

If $e\in T$, its single-edge route is the unique tree graph path from $u$ to $v$, so $j_T(u,v)=1$. If $e\notin T$, that tree graph path does not use $e$, so $j_T(u,v)=0$. ThereforeBy Ohm's law, $J(u,v)=h(u)-h(v)$; for a unit terminal current, this voltage difference is exactly the effective resistance between the terminals. We have proved the requested edge-inclusion formula for a uniform spanning tree:Now use linearity of expectation and the fact that every spanning tree has exactly $n-1$ edges. This gives Foster's theorem:The sums here use unoriented edges exactly once. For unit conductance this is the displayed identity; with conductances $c_e$ the inclusion probability and summand become $c_e{R}_{\mathrm{eff}}(e)$.

Use simple random walk on the loopless undirected unweighted graph, with transition probability $1/\mathrm{deg}(x)$ to each graph neighbour, and take $a\ne z$. Set $T_z=\inf \{t\ge 0:X_t=z\}$. Returns to $a$ before visiting $z$ do not end the commute. After its first visit to $z$, stop at the subsequent first visit to $a$, so $\tau =T_z+T_a\circ {\vartheta }_{T_z}$. Count transitions with departure times $0\le t<\tau$, including the final step arriving at $a$. The expected hitting times are finite: in a finite connected graph there is a uniformly positive probability to reach any fixed target within a fixed number of steps, giving a geometric tail bound.

We prove the needed killed-walk occupation voltage identity from its visit balance equation. DefineFor $x\ne z$, counting arrivals before absorption givesHence, for the Graph Laplacian $Lv(x)=\mathrm{deg}(x)v(x)-{\sum }_{y\sim x}v(y)$, we have $Lv(x)=1_{\{x=a\}}$ away from $z$. Since the sum of all Laplacian coordinates is zero, $Lv(z)=-1$. Thus $v$ is the electrical voltage of a unit flow from $a$ to $z$, grounded at $z$, and $v(a)=R_{\mathrm{eff}}(a,z)$.

Construct $w(x)=g_{za}(x)/\mathrm{deg}(x)$ for the opposite killed leg. It satisfies $Lw={\delta }_z-{\delta }_a$ and $w(a)=0$. Therefore $L(v+w)=0$. The harmonic maximum principle on a finite graph makes $v+w$ constant: at a maximum its value equals the average over its graph neighbours, forcing all of them to share that value, and connectedness propagates it. At $a$ the constant equals $R_{\mathrm{eff}}(a,z)$.

For any oriented edge $(x,y)$, each visit to $x$ before the target hit produces a transition to $y$ with conditional probability $1/\mathrm{deg}(x)$. The expected traversals during the first killed leg are therefore $v(x)$, and those during the second leg are $w(x)$ by the Strong Markov property at $T_z$. This proves directed edge occupation in a random-walk commute:In particular, both directions of every unoriented edge have this same expected value, although their counts on an individual commute need not coincide. Summing over the $2|E|$ directed edges also recovers the commute time identity ${\mathbb{E}}_a\tau =2|E|R_{\mathrm{eff}}(a,z)$.

The distinct-terminal convention is necessary for the printed “returns” interpretation. If $a=z$ and $\tau$ is the first positive return, a graph with two graph vertices and one edge gives $S(a,y)=1$ while $R_{\mathrm{eff}}(a,a)=0$. Thus either take $a\ne z$, as intended, or explicitly define the degenerate commute as $\tau =0$. A general nonuniform random walk is not covered by the unweighted simple-walk formula.

Group the random resistance cost by unoriented edges. For $e=\{x,y\}$, its resistance contributes once whenever the walk traverses it in either direction. Pathwise,The graph is finite and the commute has finite expected value. Linearity of expectation, or Tonelli theorem for these nonnegative terms, therefore permits taking expectations of the sum. By the directed edge occupation in a random-walk commute, both directed traversal counts have expected value $R_{\mathrm{eff}}(a,z)$. ThusApply Foster's theorem from part (a) to obtainThe factor two counts the two orientations of each edge. It would be lost by treating the expected directed count in part (b) as an expected count for both directions together. The same distinct-terminal and simple-random-walk conventions from part (b) remain in force.

Let $\mathcal{T}(G)$ be the finite set of spanning trees of the connected graph. A spanning tree has all the graph vertices, is connected, and contains no graph cycle. Connectedness ensures that at least one exists, for example by repeatedly deleting an edge of a graph cycle while retaining connectedness. The uniform spanning tree is the random edge set $T$ withEvery tree has $|V|-1$ edges, so uniformity is over complete trees, not over arbitrary subsets of that size. Such subsets can be disconnected or cyclic and do not belong to $\mathcal{T}(G)$. Nor are indicators of the tree edges generally independent. The matrix-tree theorem gives $|\mathcal{T}(G)|$ as any cofactor of the unweighted Graph Laplacian if a numerical normalizing constant is needed.

This definition uses the ordinary undirected unweighted graph convention. With positive conductances the analogous law for spanning trees weights $t$ proportionally to ${\prod }_{e\in t}{c}_e$ and is generally not uniform. When the graph has one graph vertex the empty edge set is the unique spanning tree, so its probability is one.

For finite $G$, choose a root $r$ and an ordering of the other graph vertices. Begin with a tree consisting only of $r$. At each stage take the first graph vertex not already in the tree, run an independent simple random walk from it until its first hit of the existing tree, erase the loops chronologically, and add the resulting graph path. In chronological loop erasure, revisiting a graph vertex already in the current list deletes all list entries after that graph vertex; otherwise append the new graph vertex. The retained final graph path is simple, with new internal graph vertices and exactly one endpoint in the old tree. Adding it therefore preserves connectedness and creates no graph cycle. Graph vertices already included are skipped. On a finite connected graph each walk hits the existing tree almost surely; after finitely many attachments all graph vertices are present.

The Wilson algorithm theorem states that this tree is a uniform spanning tree, independently of the chosen root and order. The required hypotheses here are an undirected connected finite graph and transition probabilities proportional to conductance; unit conductances give simple random walk and the uniform law. This is the sampling result being described, not a claim that arbitrary walk transition probabilities would produce a uniform tree.

For an infinite connected locally finite recurrent graph, choose a fixed finite root $r$ and enumerate all graph vertices. A connected locally finite graph is countable, since its finite-radius balls are finite. Run the same algorithm successively. Recurrence and irreducibility imply that a walk from any graph vertex hits $r$ almost surely, hence hits the current tree. Every attached graph path is therefore finite. The countable intersection of these probability-one events has probability one, and the increasing union is an acyclic connected subgraph containing all graph vertices: each graph vertex receives a finite route to $r$.

This gives the uniform spanning tree of a recurrent infinite graph. It is not a uniform counting measure on all infinite spanning trees. Equivalently, its joint probability distributions on finite collections of edges are the common limits of uniform spanning trees along any graph exhaustion, using the approximations defining the free uniform spanning forest and the wired uniform spanning forest. To check that equivalence, fix a finite edge set $F$ and run the chosen enumeration through a finite prefix containing all its endpoints. Once those graph vertices are in the tree, later attachments cannot add an edge with both endpoints already included, so membership of $F$ is settled permanently. Only finitely many random walks have been used; each has a finite trajectory almost surely. All their visited graph vertices and all their graph neighbours lie in a sufficiently large exhaustion set. Couple finite-volume and infinite-volume walks with the same steps there. Whether the exterior is deleted or wired to one boundary graph vertex, these initial trajectories and their loop erasures then agree. This proves convergence of the probabilities of cylinder sets determined by those edges.

The limiting law is unchanged by the root and ordering because the laws on finite graphs have this invariance. The same coupling shows that changing the exhaustion also leaves the law unchanged. ThusThe result is a single spanning tree because the graph is recurrent. On a transient graph, infinite-volume spanning-tree limits may instead be uniform spanning forests; recurrence cannot be omitted from this conclusion.

First take $x\ne y$. Sample a uniform spanning tree using Wilson's algorithm rooted at $y$, with $x$ processed first. The first walk runs from $x$ until its first hit of $y$, so its loop erasure is exactly the unique $x$-to-$y$ graph path of the resulting tree. Later attachments cannot alter that graph path. This is the uniform-spanning-tree path representation of loop-erased random walk.

Instead root Wilson's algorithm at $x$ and process $y$ first. The resulting tree still has the same uniform law, while its first attachment has the law of loop-erased random walk from $y$ to $x$. A tree's graph path between two fixed graph vertices has the same unoriented edge set in both directions. Pushing the common tree law through this graph path map therefore givesHere $\mathcal{E}$ means the set of unoriented edges. In fact, reversing the complete graph vertex sequence from the first loop-erased walk has the law of the second, giving reversibility of loop-erased random walk. If $x=y$ and first hitting allows time zero, both graph paths are empty and the equality is immediate.

This is a distributional assertion, not a pathwise identity between chronological loop erasure and reversal. For example, the valid walk trajectory $x,a,b,a,c,b,y$ in a graph containing the triangle on $a,b,c$ and the edges $xa,by$ erases forward to $x,a,c,b,y$. Erasing the reversed trajectory and then reversing the result gives $x,a,b,y$, a different graph path. Wilson's root-independent tree law is what proves the stated equality of probability distributions.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

The symmetric group has $n!$ elements, not the printed order $n$. For $n\ge 2$ the identity and the $n-1$ adjacent transpositions have probabilities $1/n$, so the specified transition matrix is normalized. Its stationary distribution is uniform and it satisfies detailed balance, since each transposition is its own inverse. Adjacent transpositions generate the symmetric group, making this an irreducible Markov chain; the positive holding probability makes it an aperiodic Markov chain.

For a fixed label $i$, right multiplication swaps positions, so $f(\sigma )={\sigma }^{-1}(i)$ is the label's position. Under the uniform distribution on a finite set it is uniform on $\{1,\dots ,n\}$. ThereforeFor each generator $s_j=(j,j+1)$, the squared change $[f(\sigma s_j)-f(\sigma ){]}^2$ is one if the label occupies either of the two positions, and zero otherwise. Its stationary expected value is $2/n$, givingThus the tagged-label Rayleigh quotient for adjacent transpositions gives the corrected bound

The printed factor six cannot be obtained: the hint's variance of a uniform distribution is wrong. Directly, $\mathrm{Var}(U)={\int }_0^1{u}^{2}du-({\int }_0^{1}udu{)}^2=1/12$. Bounded convergence in distribution of $U_n$ does imply convergence of its first two moments, hence convergence of its variance, but the limit is $1/12$.

More strongly, the claimed asymptotic bound itself is false for this kernel. The Aldous spectral gap theorem states that the interchange process on a finite connected weighted graph has the same spectral gap as the single-label random walk with identical edge rates; see Theorem 1.1 and its proof. Here $P-I$ is that interchange process generator on a path graph with edge rates $1/n$. For the tagged-label generator, direct substitution of $g_{\ell }(k)=\cos (\ell \pi (k-1/2)/n)$ gives eigenvalues $-2(1-\cos (\ell \pi /n))/n$, $0\le \ell <n$, including the reflecting endpoint equations. These $n$ distinct eigenvalues exhaust the tagged-label function space. Consequently the theorem yieldswhich exceeds $6/n^3$ asymptotically. This use of a substantial external theorem is only to certify the printed claim's failure; the corrected factor-twelve upper bound above follows directly from the requested test function and the variational formula.

Let $v_1\cdots v_r$ be a longest path in the finite simple graph. Every neighbour of either endpoint lies on it. Among indices $1\le i<r$, consider those with edges $v_1{v}_{i+1}$ and those with edges $v_i{v}_r$. Their cardinalities are the endpoint degrees, whose sum is at least $n\ge r$, exceeding the $r-1$ available indices. They intersect, producing the cycleEvery connected component has at least $\delta (G)+1>n/2$ vertices, so the graph is connected. If $r<n$, an edge joins a vertex outside this cycle to it, and breaking the cycle there gives a longer path, a contradiction. Hence $r=n$ and this is a Hamilton cycle, proving Dirac theorem.

The complete graph $K_4$ is planar, has chromatic number four and a Hamilton cycle. Adding one leaf attached to one of its vertices preserves planarity and chromatic number four but prevents a Hamilton cycle, since a leaf has degree one.

For the last claim, the given plane embedding has all vertices on the outer cycle. Add noncrossing diagonals until its polygonal interior is triangulated. A triangulated polygon has a vertex incident with only its two boundary neighbours: for example the dual graph of its triangles is a tree and a leaf triangle supplies an ear. Remove this vertex. Inductively three-colour the smaller triangulated polygon, starting with a triangle. The removed vertex's neighbours are adjacent and have different colours, so it can receive the third. Restriction to the original graph givesThis uses the specified embedding, not the false assertion that every planar Hamiltonian graph is three-colourable; $K_4$ is already a counterexample to that broader assertion.

For a finite bipartite graph with left vertex set $L$, Hall marriage theorem says that a matching in a graph covering $L$ exists exactly when $|N(S)|\ge |S|$ for every $S\subseteq L$. Necessity follows because distinct matched neighbors of $S$ lie in $N(S)$.

For sufficiency, induct on $|L|$, the empty case being immediate. If a nonempty proper subset $S$ has $|N(S)|=|S|$, apply the induction hypothesis to the graph on $S,N(S)$. For the remaining graph, every $T\subseteq L\setminus S$ satisfiesso induction matches the remaining vertices too. If there is no such tight subset and $|L|>1$, every nonempty proper subset has at least one extra neighbor. Select an edge and remove its endpoints. Each subset of the remaining left vertices loses at most one neighbor, so it still satisfies the required condition; induction supplies the rest of the matching in a graph. The case $|L|=1$ follows directly from the condition.

For a doubly stochastic matrix $A=(a_{ij})$, make a bipartite graph joining row $i$ to column $j$ when $a_{ij}>0$. For a row subset $S$,Hall marriage theorem therefore gives a perfect matching, or a permutation $\sigma$, with $\boxed{a_{i,\sigma (i)}>0\text{ for every }i}$.

No, it cannot deformation retract onto a graph. Each sphere contributes a degree-$2$ homology class independent of the other two. Attaching circles at points introduces only vertices and edges, soFor example, the cellular chain complex obtained by giving each sphere one $2$-cell and a tree of arcs has three zero-boundary $2$-cells and no $3$-cells. Every graph has zero degree-$2$ homology. A deformation retraction would induce a homotopy equivalence and an isomorphism on homology, which is impossible. An ordinary retraction has no such requirement.

Yes, it retracts onto a graph. On each of the three spheres, choose an arc joining the two marked points. A sphere retracts onto this arc: identify it with the vertical meridian parameterized by the height coordinate, and send every point of a given height to the meridian point at that height. This map is continuous and fixes the meridian. Apply these three retractions and fix the lifted circles pointwise; the maps agree at all attachment points. The resulting connected graph has six vertices and nine edges, so its fundamental group is free of rank $\boxed{9-6+1=4}$.