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A simply connected space is path connected and has trivial fundamental group: every closed loop contracts to a point. The three-sphere is simply connected, whereas Real projective space ${\mathbb{RP}}^3$ has fundamental group ${\mathbb{Z}}_2$.

For positive $p,q$, quotient $S^p\times S^q$ by $(x,y)\mapsto (-x,-y)$. This is a free double covering space action. Its closed manifold quotient is orientable exactly when $p+q$ is even, because the product mapping degree is $(-1{)}^{p+q+2}$. If $p,q\ge 2$, the product is simply connected and the quotient has fundamental group $\mathbb{Z}/2$. For $p=q=2$, the quotient has integral cohomology groups $\mathbb{Z}$ in degrees $0,4$, $\mathbb{Z}/2$ in degrees $2,3$, and zero elsewhere. Every positive-degree cup product is zero. These conclusions follow from Poincare duality, the Euler characteristic under a finite covering, and the universal coefficient theorem for cohomology.

The Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$. Under SU(2) as the three-sphere, multiplication by $-I$ is the antipodal map, so the quotient is Real projective space ${\mathbb{RP}}^3$. Equivalently, an axis-angle ball of radius $\pi$ has opposite boundary points identified. The fundamental group is ${\mathbb{Z}}_2$.

Every connected closed orientable three-manifold with finite fundamental group admits spherical geometry. With cyclic fundamental group, it is a lens space.

A topological invariant is a property or mathematical object preserved by homeomorphisms of topological spaces. Connectedness and compactness are examples. The fundamental group is an invariant up to group isomorphism; it is also preserved up to isomorphism by homotopy equivalence, a stronger invariance property which does not hold for every topological invariant.

A topological $n$-manifold without boundary is a topological space $M$ satisfying three conditions: it is Hausdorff, it is second countable, and every point $p\in M$ lies in an open set $U\subseteq M$ with a homeomorphism $\phi :U\to V$ onto an open set $V\subseteq {\mathbb{R}}^n$. Here $n\ge 0$ is a fixed dimension of a manifold, and ${\mathbb{R}}^n$ has the usual topology of Euclidean space. The pair $(U,\phi )$ is a coordinate chart. Unless explicitly qualified, “topological manifold” here means without boundary.^[ref]

The Hausdorff condition requires disjoint open neighbourhoods for distinct points. Second countability requires a countable basis of a topology. Neither follows merely from the local homeomorphisms: non-Hausdorff manifolds fail the first condition, while an uncountable disjoint union of copies of the real line fails the second. Indeed, a basis of a topology must supply a different nonempty member inside each copy, so it cannot be countable.

The circle and the real line are both one-dimensional topological manifolds, with the same local model. Their global topologies differ: the circle is compact, while the real line is not. By contrast, the junction of a topological tripod already fails the local test. Thus local coordinate charts specify what kind of small pieces are allowed; global topological invariants distinguish how those pieces fit together.^[ref]

The usefulness of a topological manifold is that local questions about continuity can be transported into Euclidean space, while compactness, connectedness, and the fundamental group still record global information. For calculus, further structure is needed: a smooth atlas requires smooth transition maps, letting the chain rule reconcile derivatives computed in different coordinate charts. A topological atlas by itself only guarantees homeomorphisms between coordinates.^[ref]

Consider a connected assembly of $n$ closed discs formed by identifying $e$ pairs of boundary arcs, each arc used once and all arcs in a given tile having disjoint endpoints. Its gluing graph has one vertex per tile and one edge per paired arc. A spanning tree joins the initial discs into a single closed disc; each of the remaining $e-n+1$ pairings attaches a band. This is a thickening of the gluing graph, so it is homotopy equivalent to that graph. Thus its Euler characteristic is $n-e$ and its fundamental group is the free group $F_{e-n+1}$. If the gluing graph is a tree, the assembled surface is itself a closed disc.

Suppose each pairing preserves the positive boundary parameter of the reference discs. A global orientation must then assign opposite orientation signs to adjacent tiles, because their induced boundary orientations on an interior seam must oppose one another. Such signs exist exactly when the gluing graph is bipartite. An odd gluing cycle therefore proves that the assembly is a nonorientable surface. A Dirichlet boundary condition or a Neumann boundary condition on an unpaired arc has no effect on this topological calculation.

A triangular group presentation has relators of word length at most three. Every finitely presented group admits one. For a relator $s_1\cdots s_{\ell }$ of length $\ell \ge 4$, introduce generators $y_2,\dots ,y_{\ell -2}$ and replace it byThe defining relations say $y_j=s_1\cdots s_j$; eliminating those generators by Tietze transformations recovers the original relation. Every new relator has length three. Geometrically these operations triangulate each relator disk in its presentation complex by inserting diagonals, so they preserve the fundamental group.

The mapping torus of a torus map with this monodromy has fundamental group generated by $x,y,t$ with $x$ central and $ty{t}^{-1}=x^{b}y$. Normal forms $x^m{y}^n{t}^k$ multiply byThis proves torsion-freeness and nilpotence of class at most two. For $b\ne 0$ its center is $⟨x⟩$ and commutator subgroup $⟨x^b⟩$; for $b=0$ it is ${\mathbb{Z}}^3$. These realize all noncyclic torsion-free nilpotent closed orientable three-manifold groups.

A unitary representation of the fundamental group of a connected complex manifold gives a bundle $(\tilde{X}\times V)/{\pi }_{1}X$ with locally constant unitary transition matrices. Its induced flat unitary connection has $(0,1)$ part equal to its Dolbeault operator, so it is the Chern connection for the induced Hermitian metric. On a compact Riemann surface, zero curvature gives degree zero; the subbundle curvature inequality makes the bundle semistable.

For a connected space, this is the connected covering corresponding to the kernel of the homomorphism from its fundamental group to its abelianization. Its deck transformation group is $H_1(Y;\mathbb{Z})$. Lifted cells make its cellular chain complex a complex over the group ring $\mathbb{Z}[H_1(Y)]$.

The universal cover of a connected closed manifold of dimension three with finite fundamental group is a closed simply connected three-manifold. It is orientable, since its orientation character has trivial domain. Its first integral homology vanishes, and the universal coefficient theorem for cohomology then gives $H^1=0$. Integral Poincare duality gives $H_2=H^1=0$, while $H_0$ and $H_3$ are $\mathbb{Z}$. Thus it is an integral homology sphere. This proof does not require classifying simply connected three-manifolds.

The Warsaw circle joins the two path components of a closed topologist's sine curve by an arc attached at their outer endpoints. It is path-connected and has trivial fundamental group, yet admits a continuous map to the circle with no continuous lift to its real-line covering space. Set the map equal to $1$ on the closed topologist's sine curve and wind once on the joining arc. A lift would be constant on each path component of the sine curve, with the same constant on both by continuity, but would change by $1$ along the joining arc. This proves that the map is not null-homotopic, by the homotopy lifting theorem for a covering map. Thus finiteness of the fundamental group alone cannot replace local path connectedness in the usual lifting criterion for a covering space.

The wedge of two circles is a graph with one vertex and two loop edges. Its fundamental group is the free group on the two oriented loops.