Examples

A 2-dimensional hole (a hole with a 1-dimensional boundary).

• A 2-dimensional hole (a hole with a 1-dimensional boundary) is a circle (S¹) in X, that cannot be shrunk continuously to a point in X. An example is shown on the figure at the right. The yellow region is the topological space X; it is a pentagon with a triangle removed. The blue circle is a 1-dimensional sphere in X. It cannot be shrunk continuously to a point in X; therefore; X has a 2-dimensional hole. Another example is the punctured plane - the Euclidean plane with a single point removed, ${\displaystyle \mathbb {R} ^{2}\setminus \{(0,0)\}}$. To make a 2-dimensional hole in a 3-dimensional ball, make a tunnel through it.^[1] In general, a space contains a 1-dimensional-boundary hole if and only if it is not simply-connected. Hence, simply-connected is equivalent to 1-connected. X is 0-connected but not 1-connected, so ${\displaystyle {\text{conn}}_{\pi }(X)=0}$. The lowest dimension of a hole is 2, so ${\displaystyle \eta _{\pi }(X)=2}$.
  

  A 3-dimensional hole.
• A 3-dimensional hole (a hole with a 2-dimensional boundary) is shown on the figure at the right. Here, X is a cube (yellow) with a ball removed (white). The 2-dimensional sphere (blue) cannot be continuously shrunk to a single point. X is simply-connected but not 2-connected, so ${\displaystyle {\text{conn}}_{\pi }(X)=1}$. The smallest dimension of a hole is 3, so ${\displaystyle \eta _{\pi }(X)=3}$.

A 1-dimensional hole.

• For a 1-dimensional hole (a hole with a 0-dimensional boundary) we need to consider ${\displaystyle S^{0}}$ - the zero-dimensional sphere. What is a zero dimensional sphere? - For every integer d, the sphere ${\displaystyle S^{d}}$ is the boundary of the (d+1)-dimensional ball ${\displaystyle B^{d+1}}$. So ${\displaystyle S^{0}}$ is the boundary of ${\displaystyle B^{1}}$, which is the segment [0,1]. Therefore, ${\displaystyle S^{0}}$ is the set of two disjoint points {0, 1}. A zero-dimensional sphere in X is just a set of two points in X. If there is such a set, that cannot be continuously shrunk to a single point in X (or continuously extended to a segment in X), this means that there is no path between the two points, that is, X is not path-connected; see the figure at the right. Hence, path-connected is equivalent to 0-connected. X is not 0-connected, so ${\displaystyle {\text{conn}}_{\pi }(X)=-1}$. The lowest dimension of a hole is 1, so ${\displaystyle \eta _{\pi }(X)=1}$.
• A 0-dimensional hole is a missing 0-dimensional ball. A 0-dimensional ball is a single point; its boundary ${\displaystyle S^{-1}}$ is an empty set. Therefore, the existence of a 0-dimensional hole is equivalent to the space being empty. Hence, non-empty is equivalent to (-1)-connected. For an empty space X, ${\displaystyle {\text{conn}}_{\pi }(X)=-2}$ and ${\displaystyle \eta _{\pi }(X)=0}$, which is its smallest possible value.
• A ball has no holes of any dimension. Therefore, its connectivity is infinite: ${\displaystyle \eta _{\pi }(X)={\text{conn}}_{\pi }(X)=\infty }$.

Homotopical connectivity of spheres

In general, for every integer d, ${\displaystyle {\text{conn}}_{\pi }(S^{d})=d-1}$ (and ${\displaystyle \eta _{\pi }(S^{d})=d+1}$)^{[1]: 79, Thm.4.3.2} The proof requires two directions:

• Proving that ${\displaystyle {\text{conn}}_{\pi }(S^{d})<d}$, that is, ${\displaystyle S^{d}}$ cannot be continuously shrunk to a single point. This can be proved using the Borsuk–Ulam theorem.
• Proving that ${\displaystyle {\text{conn}}_{\pi }(S^{d})\geq d-1}$, that is, that is, every continuous map ${\displaystyle S^{k}\to S^{d}}$ for ${\displaystyle k<d}$ can be continuously shrunk to a single point.

Examples

• A space X is (−1)-connected if and only if it is non-empty.
• A space X is 0-connected if and only if it is non-empty and path-connected.
• A space is 1-connected if and only if it is simply connected.
• An n-sphere is (n − 1)-connected.

Interpretation

This is instructive for a subset: an n-connected inclusion ${\displaystyle A\hookrightarrow X}$ is one such that, up to dimension n − 1, homotopies in the larger space X can be homotoped into homotopies in the subset A.

For example, for an inclusion map ${\displaystyle A\hookrightarrow X}$ to be 1-connected, it must be:

• onto ${\displaystyle \pi _{0}(X),}$
• one-to-one on ${\displaystyle \pi _{0}(A)\to \pi _{0}(X),}$ and
• onto ${\displaystyle \pi _{1}(X).}$

One-to-one on ${\displaystyle \pi _{0}(A)\to \pi _{0}(X)}$ means that if there is a path connecting two points ${\displaystyle a,b\in A}$ by passing through X, there is a path in A connecting them, while onto ${\displaystyle \pi _{1}(X)}$ means that in fact a path in X is homotopic to a path in A.

In other words, a function which is an isomorphism on ${\displaystyle \pi _{n-1}(A)\to \pi _{n-1}(X)}$ only implies that any elements of ${\displaystyle \pi _{n-1}(A)}$ that are homotopic in X are abstractly homotopic in A – the homotopy in A may be unrelated to the homotopy in X – while being n-connected (so also onto ${\displaystyle \pi _{n}(X)}$) means that (up to dimension n − 1) homotopies in X can be pushed into homotopies in A.

This gives a more concrete explanation for the utility of the definition of n-connectedness: for example, a space where the inclusion of the k-skeleton is n-connected (for n > k) – such as the inclusion of a point in the n-sphere – has the property that any cells in dimensions between k and n do not affect the lower-dimensional homotopy types.

Homology

Hurewicz theorem relates the homotopical connectivity ${\displaystyle {\text{conn}}_{\pi }(X)}$ to the homological connectivity, denoted by ${\displaystyle {\text{conn}}_{H}(X)}$. This is useful for computing homotopical connectivity, since the homological groups can be computed more easily.

Suppose first that X is simply-connected, that is, ${\displaystyle {\text{conn}}_{\pi }(X)\geq 1}$. Let ${\displaystyle n:={\text{conn}}_{\pi }(X)+1\geq 2}$; so ${\displaystyle \pi _{i}(X)=0}$ for all ${\displaystyle i<n}$, and ${\displaystyle \pi _{n}(X)\neq 0}$. Hurewicz theorem^{[5]: 366, Thm.4.32} says that, in this case, ${\displaystyle {\tilde {H_{i}}}(X)=0}$ for all ${\displaystyle i<n}$, and ${\displaystyle {\tilde {H_{n}}}(X)}$ is isomorphic to ${\displaystyle \pi _{n}(X)}$, so ${\displaystyle {\tilde {H_{n}}}(X)\neq 0}$ too. Therefore:

$${\displaystyle {\text{conn}}_{H}(X)={\text{conn}}_{\pi }(X).}$$

If X is not simply-connected (${\displaystyle {\text{conn}}_{\pi }(X)\leq 0}$), then

$${\displaystyle {\text{conn}}_{H}(X)\geq {\text{conn}}_{\pi }(X)}$$

still holds. When ${\displaystyle {\text{conn}}_{\pi }(X)\leq -1}$ this is trivial. When ${\displaystyle {\text{conn}}_{\pi }(X)=0}$ (so X is path-connected but not simply-connected), one should prove that ${\displaystyle {\tilde {H_{0}}}(X)=0}$.

The inequality may be strict: there are spaces in which ${\displaystyle {\text{conn}}_{\pi }(X)=0}$ but ${\displaystyle {\text{conn}}_{H}(X)=\infty }$.^[6]

By definition, the k-th homology group of a simplicial complex depends only on the simplices of dimension at most k+1 (see simplicial homology). Therefore, the above theorem implies that a simplicial complex K is k-connected if and only if its (k+1)-dimensional skeleton (the subset of K containing only simplices of dimension at most k+1) is k-connected.:^{[1]: 80, Prop.4.4.2}

Join

Let K and L be non-empty cell complexes. Their join is commonly denoted by ${\displaystyle K*L}$. Then:^{[1]: 81, Prop.4.4.3}

$${\displaystyle {\text{conn}}_{\pi }(K*L)\geq {\text{conn}}_{\pi }(K)+{\text{conn}}_{\pi }(L)+2.}$$

The identity is simpler with the eta notation:

$${\displaystyle \eta _{\pi }(K*L)\geq \eta _{\pi }(K)+\eta _{\pi }(L).}$$

As an example, let ${\displaystyle K=L=S^{0}=}$ a set of two disconnected points. There is a 1-dimensional hole between the points, so the eta is 1. The join ${\displaystyle K*L}$ is a square, which is homeomorphic to a circle, so its eta is 2. The join of this square with a third copy of K is a octahedron, which is homeomorphic to ${\displaystyle S^{2}}$, and its eta is 3. In general, the join of n copies of ${\displaystyle S^{0}}$ is homeomorphic to ${\displaystyle S^{n-1}}$ and its eta is n.

The general proof is based on a similar formula for the homological connectivity.

Nerve

Let K₁,...,K_n be abstract simplicial complexes, and denote their union by K.

Denote the nerve complex of {K₁, ... , K_n} (the abstract complex recording the intersection pattern of the K_i) by N.

If, for each nonempty ${\displaystyle J\subset I}$, the intersection ${\textstyle \bigcap _{i\in J}U_{i}}$ is either empty or (k−|J|+1)-connected, then for every j ≤ k, the j-th homotopy group of N is isomorphic to the j-th homotopy group of K.

In particular, N is k-connected if-and-only-if K is k-connected.^[7]: Thm.6