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A Serre fibration has the homotopy lifting property for disks: for every $k\ge 0$, a map $D^k\to E$ and a homotopy $D^k\times I\to B$ starting at its composite with $p:E\to B$ admit a compatible lift $D^k\times I\to E$. Equivalently, it has that lifting property for CW complexes. Relative lifting for CW pairs follows by attaching cells.

Let $i:F\hookrightarrow E$ be the inclusion of the chosen fiber. The long exact sequence of homotopy groups of a fibration isIts low-dimensional end isThe maps $i_{\ast }$ and $p_{\ast }$ are induced by inclusion and projection on based maps; at degree zero they send a path component to its containing or image component. The last part is an exact sequence of pointed sets, not in general a sequence of groups.

For $\partial :{\pi }_q(B)\to {\pi }_{q-1}(F)$, represent a class by a based cube $a:I^{q-1}\times I\to B$, constant at $b_0$ on its boundary. Lift it from the constant map $e_0$ on the bottom and side faces, using relative disk lifting. Its top face lands in $F$ and is constant on that face's boundary; its class is $\partial [a]$. The choice of lifts does not change the resulting class. For $q=1$, lift a based loop starting at $e_0$ and take the path component of its endpoint in $F$. This fixes the boundary-map convention and defines every map in the displayed sequence.

For the splitting assertion, first take $n\ge 1$. We use the cohomological Serre spectral sequence with coefficient group $G$:Since $B$ is simply connected, these coefficient systems are constant. The Eilenberg–MacLane space $F=K(G,n)$ has $H^t(F;G)=0$ for $0<t<n$, andFor $n=1$ this uses $H_1(F)=G$, with $G$ abelian; for $n>1$ it follows from the Hurewicz theorem and the universal coefficient theorem for cohomology. Let $u$ correspond to ${\mathrm{id}}_G$, the universal cohomology class of an Eilenberg–MacLane space.

In bidegree $(0,n)$ there are no incoming differentials. For $2\le r\le n$, an outgoing differential lands in a vanishing fiber-cohomology row. The only remaining possible differential isThus $u$ survives. The edge map, which is restriction to $F$, supplies a classBy representability of cohomology by Eilenberg–MacLane spaces, choose a based map $q:E\to K(G,n)$ representing $c$. On $F$, $q$ induces the identity on ${\pi }_n=G$ and is an isomorphism on every homotopy group, since the other positive groups vanish.

Consider $h=(p,q):E\to B\times K(G,n)$. This is a map of fibrations over $B$, whose map on the fiber is the weak equivalence just obtained. The map on the base is the identity. Comparing the two instances of the long exact sequence of homotopy groups of a fibration and applying exactness, or the Five lemma in the group-valued range, shows that $h$ induces all homotopy isomorphisms. The spaces are path-connected because the base and fiber are. Hence the splitting of a simply connected Eilenberg–MacLane fibration givesThe crucial use of $H^{n+1}(B;G)=0$ is to lift the fiber's identity class; existence of a section alone was not assumed to prove a product splitting.

If $K(G,0)$ is allowed, its components have trivial positive homotopy groups. The simply connected base has no monodromy on this discrete set. Obstruction theory on the CW base gives a section in each fiber component, since all positive-dimensional fiber obstructions vanish and the component system is constant. Together these sections give $B\times G\to E$, a weak equivalence on each component. Thus the same conclusion also covers that interpretation.

A Serre fibration $p:E\to B$ has the homotopy lifting property for every disc: given $h:D^k\times I\to B$ and ${\tilde{h}}_0:D^k\to E$ with $p{\tilde{h}}_0=h(-,0)$, there is a lift $\tilde{h}$ extending ${\tilde{h}}_0$. Equivalently one may test this property on CW complexes.

For a map $g:E\to B$, its mapping-path fibration replacement isPath spaces have the compact-open topology, with the usual compactly generated convention. The inclusion $s(e)=(e,{\mathrm{const}}_{g(e)})$ satisfies $ps=g$. Forgetting the path is a homotopy inverse to $s$: shorten $\gamma$ to $u\mapsto \gamma (tu)$, from its constant initial path at $t=0$ to the original path at $t=1$. The equivalence here means a commuting triangle over the same base whose map of total spaces is a homotopy equivalence; it does not assert a fibrewise homotopy inverse over the base. Appending the successive segments of a moving base path gives the homotopy lifting property for $p$, even for arbitrary parameter spaces, so $p$ is in particular a Serre fibration. Its ordinary fibre over $b_0$ is the homotopy fibre

For a based Serre fibration, with fibre inclusion $j:F\hookrightarrow E$, the long exact sequence of homotopy groups of a fibration isThe maps $j_{\ast },p_{\ast }$ compose based sphere representatives with $j,p$; the maps on path components do the same with points. To define $\partial$, represent a class by $(I^k,\partial I^k)\to (B,b_0)$ and lift the cube starting at $e_0$ on its bottom face, keeping its vertical sides at $e_0$. The opposite face then gives $(I^{k-1},\partial I^{k-1})\to (F,e_0)$. Relative lifting for a CW pair gives this construction; its based homotopy class is independent of choices. At $k=1$ it records the component of the endpoint of the lifted loop. The last terms are pointed sets rather than groups in general.

Now write $W=S^1\vee S^n$ and let $f:W\to S^n$ be the collapse map. Inclusion of the $n$-sphere is a section, so $f_{\ast }:{\pi }_k(W)\to {\pi }_k(S^n)$ is surjective in every positive degree. Thus all connecting maps in its homotopy fibre sequence vanish. The fibre is path-connected, andThe section also gives the first assertion when $n=2$, by killing the connecting map from ${\pi }_2(S^2)$.

The universal cover of $W$ consists of a line with an $n$-sphere attached at each integer. It is homotopy equivalent to the wedge sum ${\bigvee }_{k\in \mathbb{Z}}{S}^n$. This cover is $(n-1)$-connected, so the Hurewicz theorem identifies its $n$th homotopy group with its $n$th homology. Choose a generator $t$ of the fundamental group whose action shifts the sphere indexed by $k$ to that indexed by $k+1$. The resulting homotopy groups as modules over the fundamental group giveEvery lifted sphere maps to $S^n$ with degree one. Consequently $f_{\ast }$ is the augmentation $\epsilon (\sum a_k{t}^k)=\sum a_k$, with the trivial $R$-action on $\mathbb{Z}$. The inclusion of the fibre respects the fundamental group actions, soThe augmentation ideal of the infinite cyclic group is generated by $t-1$: multiply a Laurent polynomial by a power of $t$ and use divisibility by $t-1$ when its value at one vanishes. Multiplication by $t-1$ is injective, giving the last isomorphism. This is a free module of rank one, whose underlying abelian group has countably infinite rank; its ${\pi }_1$-action is not trivial.

The given homeomorphisms and homotopy invariance of homology givebecause $n\ge 2$, while $M\simeq S^{n-1}$ is path connected and $L$ is the disjoint union of two copies of $S^{n-1}$. HenceThe degree-one and degree-zero part of the Mayer–Vietoris sequence is thereforeTaking the rank of an abelian group throughout this short exact sequence givesso $\mathrm{rank}{H}_0(N)=2$. Since the zeroth homology group is freely generated by the path components,

A topological space is connected when it is not the union of two disjoint nonempty open subsets. It is path-connected when every two points $x,y$ can be joined by a path, meaning a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.

To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. LetWhichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the supremum or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.

If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$, contradicting the connectedness just proved. Hence

Now let $X\subseteq {\mathbb{R}}^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an open ball is path-connected by straight line segments. It follows that every path component of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.

The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is locally path-connected, its path components are open, and connectedness forces there to be only one.

For the final example, putThe set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the closure of $Y$, because $(1/n,y)\to (0,y)$ for every $y\in [1/2,1]$. A connected set together with a connected subset of its closure is connected, so

It is not path-connected. The image of the first coordinate of any path in $X$ lies inbut, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup \{1/n:n\ge 1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\le y<1/2$. Hence no path joins $B$ to $Y$, and

The zeroth homology group is the free abelian group on the path components of $X$. In particular, if $X$ has finitely many path components, then