Point-set
(A compact subset need not be closed)
In the line with two origins, \([-1,1]\) together with one of the two origins is compact and is not closed, since the other origin lies in its closure. In a Hausdorff space, every compact subset is closed.
(A closed and bounded set need not be compact)
The closed unit ball of an infinite-dimensional normed space is closed and bounded, and it is not compact. By the Heine–Borel theorem, closed and bounded subsets of \(\RR^n\) are compact.
(A continuous image of a closed set need not be closed)
The projection \(\RR^2\to\RR\), \((x,y)\mapsto x\), sends the closed hyperbola \(\ts{xy=1}\) onto \(\RR\sm\ts0\), which is not closed. Continuous images of compact sets are compact, and continuous images of connected sets are connected.
(A quotient of a Hausdorff space need not be Hausdorff)
The line with two origins is the quotient of two copies of \(\RR\) obtained by identifying \(x\) in the first copy with \(x\) in the second for every \(x\neq 0\). Each copy of \(\RR\) is Hausdorff, and the two origins have no disjoint neighborhoods in the quotient.
(A connected space need not be path connected)
The topologist’s sine curve \(\ts{(x,\sin(1/x)) \st 0<x\leq 1}\union \ts{0}\times[-1,1]\) is connected and not path connected. It is also not locally connected.
(A product of connected spaces need not be connected in the box topology)
\(\RR^\omega\) is connected in the product topology and not connected in the box topology: the set of bounded sequences is open and closed in the box topology. An arbitrary product of connected spaces is connected in the product topology.
(Completeness is not a topological property)
\((0,1)\) and \(\RR\) with their standard metrics are homeomorphic, \(\RR\) is complete, and \((0,1)\) is not.
Fundamental group and covering spaces
(The fundamental group does not determine the homotopy type)
\(S^1\vee S^2\) and \(S^1\) both have fundamental group \(\ZZ\), and \(H_2(S^1\vee S^2)\cong\ZZ\) while \(H_2(S^1)=0\).
(A simply connected space need not be contractible)
\(S^2\) is simply connected, and \(H_2(S^2)\cong\ZZ\) shows that it is not contractible.
(A space need not have a universal cover)
The Hawaiian earring is path connected and locally path connected, and it is not semilocally simply connected, so it has no universal cover.
For a path-connected, locally path-connected, semilocally simply connected space \(X\) with basepoint \(x_0\), conjugacy classes of subgroups of \(\pi_1(X,x_0)\) correspond to isomorphism classes of connected covering spaces of \(X\). For the Hawaiian earring the trivial subgroup corresponds to no covering space.
Homology
(Homology does not determine the homeomorphism type)
The Poincaré homology sphere has the homology of \(S^3\), and its fundamental group is the binary icosahedral group of order \(120\), so it is not homeomorphic to \(S^3\).
(Homology does not determine the fundamental group)
For path-connected \(X\), \(H_1(X)\) is the abelianization of \(\pi_1(X)\). The fundamental group of the Poincaré homology sphere is perfect and nontrivial, so \(H_1 = 0\) while \(\pi_1\neq 1\).
(A homology isomorphism need not be a homotopy equivalence)
Collapsing the complement of an open \(3\)-ball in the Poincaré homology sphere \(P\) gives a map \(P\to S^3\) of degree \(1\), which induces isomorphisms on all homology groups. It is not a homotopy equivalence, since \(\pi_1(P)\neq 1\). By Whitehead’s theorem, a map between simply connected CW complexes inducing isomorphisms on all homology groups is a homotopy equivalence.
(Singular homology does not commute with infinite products)
Let \(X=\prod_{n\geq 1}\ts{0,1}\) be a countable product of two-point discrete spaces, the Cantor set. Its path components are its points, so \(H_0(X)\) is free abelian of uncountable rank, whereas \(\prod_{n\geq1} H_0(\ts{0,1}) \cong \prod_{n\geq 1}\ZZ^2\) is not free by the Baer–Specker theorem.
For good pointed spaces \(X_\alpha\), \(\tilde H_*(\bigvee_\alpha X_\alpha)\cong\bigoplus_\alpha \tilde H_*(X_\alpha)\).
Manifolds
(A closed manifold need not have top homology $\ZZ$)
For a connected closed \(n\)-manifold \(M\), \(H_n(M)\cong\ZZ\) if \(M\) is orientable and \(H_n(M)=0\) otherwise. The Klein bottle is non-orientable and has \(H_2=0\), while the torus has \(H_2\cong\ZZ\).
(A manifold need not be triangulable)
Every topological manifold of dimension at most \(3\) is triangulable. Freedman’s \(E_8\) manifold is a closed topological \(4\)-manifold that admits no triangulation.
(Homotopy equivalent manifolds need not be homeomorphic)
The lens spaces \(L(7,1)\) and \(L(7,2)\) are homotopy equivalent closed \(3\)-manifolds that are not homeomorphic.