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A map $p:E\to B$ is a Serre fibration if every homotopy $H:D^k\times I\to B$, $k\ge 0$, with a lift of its initial slice has a lift on the whole cylinder extending that initial lift. By attaching cells, this gives the relative homotopy lifting property for CW pairs: a specified lift on the initial slice and on the subcomplex cylinder can be extended.

Give $PY$ the compact-open topology. Suppose the initial lifting is a family of paths ${\gamma }_x$ starting at $y_0$, with ${\gamma }_x(1)=H(x,0)$. DefineThe two expressions coincide on their common boundary, the initial path at $s=0$ is exactly ${\gamma }_x$, and the endpoint is $H(x,s)$. The adjoint map on $D^k\times I\times I$ is continuous by pasting; compactness of the path parameter gives continuity into the path space. Thus endpoint evaluation $PY\to Y$ is a Serre fibration. The variable subdivision avoids changing the initial lift by an unwanted reparametrization.

For a based pair $(E,F,e_0)$, the long exact sequence of relative homotopy groups isending in ${\pi }_1(E)\to {\pi }_1(E,F)\to {\pi }_0(F)\to {\pi }_0(E)$. The low-dimensional end is an exact sequence of pointed sets with its usual actions, rather than an entirely abelian sequence.

The boundary sends a relative disk map to its restriction on the boundary sphere. An absolute class has constant boundary, so $\partial j_{\ast }=0$. Conversely, if a relative map has null-homotopic boundary in $F$, choose a based null-homotopy there. Attach this null-homotopy as a collar to the disk, leaving the original map on a smaller inner disk. The resulting boundary is constant at $e_0$, so it defines an absolute class in ${\pi }_n(E)$; inserting the collar is a relative homotopy, so its image is the original relative class. This proves $\mathrm{im}{j}_{\ast }=\ker \partial$. At $n=1$, append a path in $F$ from the terminal point to $e_0$, giving the corresponding based loop and pointed-set exactness.

Now suppose $F=p^{-1}(b_0)$. Use a cube model in which the bottom and side faces are based and the top face maps into $F$. A based $n$-cube in $B$ lifts from its constant bottom face, with sides held at $e_0$ by relative lifting. Its top face lies in $F$, proving surjectivity of the relative homotopy of a Serre fibration comparison $p_{\ast }:{\pi }_n(E,F)\to {\pi }_n(B,b_0)$.

For injectivity, lift a based homotopy of two projected cubes from the first relative lift; this ends at some lift of the second base cube. Any two lifts of that same base cube are relatively homotopic: insert an extra parameter between them, prescribe the two liftings on its two ends and the constant lift on the bottom, and apply relative lifting in the remaining cube direction. The top face stays in $F$ because its projection is $b_0$. Combining these homotopies proves injectivity. ThereforeAt $n=1$ the same construction gives a bijection of pointed sets, and one can transport the base fundamental-group law across it. An arbitrary pair's relative ${\pi }_1$ has no general group law. The printed target index $a$ is a typographical error: the map preserves degree $n$.

The last-column map $U(n)\to S^{2n-1}$ is surjective because a unit vector can be completed to a unitary basis. Its stabilizer is the block subgroup $\mathrm{diag}(U(n-1),1)$, so it induces a continuous bijection $U(n)/U(n-1)\to S^{2n-1}$. Unitary matrices form a compact set in their matrix-space topology; the quotient is compact and the sphere Hausdorff. HenceThis quotient is a fiber bundle. Near the last standard unit vector, project the other standard vectors by $I-v{v}^{\ast }$ onto $v^{\perp }$ and apply Gram-Schmidt orthogonalization; their independence persists in a small neighborhood. The resulting orthonormal columns together with $v$ give a continuous local section. Translating these sections covers the sphere and trivializes the projection, so it is a Serre fibration.

Apply this to $U(n)\to U(n+1)\to S^{2n+1}$. Its homotopy sequence containsFor $i\le 2n-1$, both outer groups vanish by cellular approximation theorem. Thus ${\pi }_i(U(n))\cong {\pi }_i(U(n+1))$ in the stated range. The component assertion also holds: diagonalizing a unitary matrix and varying its eigenvalue phases continuously connects it to the identity. This establishes the unitary sphere quotient and stability range.

The cellular approximation theorem says that a continuous map between CW complexes is homotopic to a cellular map, meaning that it sends each $q$-skeleton into the target's $q$-skeleton. Give $S^n$, for $n>0$, its CW complex structure with one zero-cell and one $n$-cell. Its $n$-skeleton is the whole sphere, so the theorem applies and gives the requested result. For $n=0$ use the two zero-cells of $S^0$. In particular the conclusion isThere is no local-finiteness assumption on $X$.

For completeness, here is a direct proof of the needed case of cellular approximation, including the point-avoidance step. We use three standard facts, stated explicitly. First, a compact subset of a CW complex lies in a finite subcomplex. Second, a continuous map from a smooth manifold into Euclidean space has arbitrarily close smooth approximations, and smooth cutoffs exist around compact subsets of open sets. Third, the Sard theorem says the critical values of a smooth map have measure zero; when the domain dimension is smaller than the target dimension, its entire image has measure zero.

The first fact has a useful short proof. If a compact set met infinitely many open cells, choose one point in each of countably many distinct cells. Closure-finiteness makes every closed cell meet the chosen set, and every subset of it, in finitely many points. The weak topology of a CW complex then makes all those subsets closed. The chosen set is an infinite closed discrete subset of a compact space, a contradiction. Taking the closures of the finitely many cells met proves the claim.

Now $f(S^n)$ lies in a finite subcomplex $Y$. Take a top-dimensional open cell $e^d$ of $Y$ with $d>n$, and identify its interior with the open unit ball in ${\mathbb{R}}^d$ using its characteristic map. On the open set $U=f^{-1}(e^d)$, write $F:U\to B^d$ for the coordinate expression. Choose $0<r^{\prime }<r^{\prime \prime }<1$ and a smooth cutoff $\chi$ supported in $F^{-1}(B_{r^{\prime \prime }})$ and equal to one near the compact set $F^{-1}({\overline{B}}_{r^{\prime }})$. By smooth approximation of maps into Euclidean space, choose a smooth $h:U\to {\mathbb{R}}^d$ with $|h-F|<\delta$, where $\delta <\min (r^{\prime }/2,1-r^{\prime \prime })$. Replace $F$ byThe straight-line interpolation is a homotopy supported away from the boundary of the cell and stays inside that cell. Where $\chi \ne 1$, we have $|F|>r^{\prime }$ and hence $|G|>r^{\prime }/2$. By the Sard theorem, choose $z\in B_{r^{\prime }/4}$ outside $h(U)$. Where $\chi =1$ we have $G=h$, and elsewhere $|G|>r^{\prime }/2$, so the modified map avoids $z$ everywhere.

Radial retraction of $D^d\setminus \{z\}$ onto its boundary fixes that boundary. It therefore descends through the characteristic map, including its possibly noninjective boundary identifications, and fixes all other cells of $Y$. Composing gives a homotopy whose endpoint misses all of $e^d$. Remove every cell of dimension greater than $n$ in descending order. There are finitely many such cells, so their homotopies concatenate to yield $g:S^n\to Y^n\subseteq X_n$. This proves dimension reduction of sphere maps into CW skeleta. The approximation step is essential: an arbitrary continuous image of $S^n$ can fill a higher-dimensional region, so dimension alone would not supply an omitted point.

The absence of $n$-simplices implies absence of every higher-dimensional simplex, since any such simplex would have an $n$-face. Therefore $X$ has a CW structure of dimension at most $n-1$. Give $S^n$ its CW structure with one zero-cell and one $n$-cell. The cellular approximation theorem states that any map of CW complexes is homotopic to a cellular map, which sends each source skeleton into the corresponding target skeleton. For $n\ge 1$, all target skeletons below dimension $n$ are the single zero-cell. The approximating map therefore sends all of $X$ there. Hence every given map is null-homotopic. This argument works for an infinite simplicial complex too; it does not require a finite subdivision or a finite collection of image simplices.

First check the signs directly. For $(x,y)\in M_i$, applying the proposed differential twice givesbecause $f$ is a chain map. Thus $M$ is a chain complex.

Let $C^{\prime }$ be the shifted chain complex with $C_i^{\prime }=C_{i-1}$ and differential $d_C$; here there is no minus sign in that differential. Inclusion in the second summand and projection onto the first give a short exact sequence of chain complexesTo compute the connecting map in its long exact sequence in homology, lift a cycle $x\in C_{i-1}$ to $(x,0)\in M_i$. Its boundary is $(0,(-1{)}^{i}f(x))$. Hence the connecting map $H_i(C^{\prime })\to H_{i-1}(D)$ is $(-1{)}^i{f}_{\ast }$, and the relevant exact portion isIf every $f_{\ast }$ is an isomorphism, exactness makes $H_i(M)=0$. Conversely, if every $H_i(M)$ is zero, the adjacent exact portions make every $f_{\ast }$ both injective and surjective. ThereforeThis is the acyclicity criterion for a mapping cone, with the degree-dependent signs adjusted to the given convention.

For the assertion about spaces, use the cellular approximation theorem to replace $f$ by a cellular map, and take the induced map of the finite free cellular chain complexes. Tensoring these complexes and their cone with ${\mathbb{F}}_p$ gives the corresponding mod-$p$ complexes and cone. The same exact-sequence argument works over ${\mathbb{F}}_p$, so the assumed homology isomorphisms implyThe universal coefficient theorem for homology givesIn particular $H_i(M)\otimes {\mathbb{F}}_p=0$ for every prime. Each $H_i(M)$ is a finitely generated abelian group. A nonzero free summand would survive tensoring with every ${\mathbb{F}}_p$, while a nonzero finite cyclic summand would survive for a prime dividing its order. Thus $H_i(M)=0$ in every degree, by detection of integral acyclicity modulo primes. Applying the cone criterion once more proves the integral homology map is an isomorphism in every degree. Finite generation is what makes detection by all prime fields sufficient.

Use the standard complex orientations and the product orientation. Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ satisfy $a^2=b^2=0$ and $⟨ab,[S^2\times S^2]⟩=1$, and let $x\in H^2({\mathbb{CP}}^2;\mathbb{Z})$ satisfy $⟨x^2,[{\mathbb{CP}}^2]⟩=1$. These are the cohomology rings of the product of two spheres and the Complex projective plane.

For a map from the Complex projective plane, write $f^{\ast }a=rx$ and $f^{\ast }b=sx$. Naturality of the cup product gives $r^2{x}^2=f^{\ast }(a^2)=0$ and $s^2{x}^2=f^{\ast }(b^2)=0$. Since $x^2$ has infinite order, $r=s=0$. It follows that $f^{\ast }(ab)=0$, so every such map has degree $0$.

In the reverse direction, write $g^{\ast }x=ra+sb$. ThenEvery even mapping degree really occurs. To construct it, identify $S^2$ with ${\mathbb{CP}}^1$ and use the Segre embeddingThe generator $x_3\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ pulls back to $a+b$: restricting to either factor gives a projective line and hence coefficient $1$. By the cellular approximation theorem, $\sigma$ is homotopic to a cellular map. Its four-dimensional domain then maps into the four-skeleton ${\mathbb{CP}}^2$ of ${\mathbb{CP}}^3$. Denote that map by $g_0$; restriction of $x_3$ to ${\mathbb{CP}}^2$ is $x$, so $g_0^{\ast }x=a+b$ and $\mathrm{deg}{g}_0=2$.

For any integer $m$, choose a map $d_m:S^2\to S^2$ of mapping degree $m$. For $m>0$ one may use $z\mapsto z^m$ on the Riemann sphere, for $m<0$ use $z\mapsto {\overline{z}}^{|m|}$, and for $m=0$ use a constant map. Then $g_m=g_0\circ (d_m\times \mathrm{id})$ satisfies $g_m^{\ast }x=ma+b$, so the possible degrees are exactly all even integers, with $g_m$ realizing $2m$. The cellular approximation here deforms this particular map into the skeleton; it does not require a retraction of ${\mathbb{CP}}^3$ onto ${\mathbb{CP}}^2$.

Finally, take the connected sum of oriented manifolds with both summands carrying their standard complex orientations. Its degree-two cohomology has generators $u,v$ withThis follows by choosing the generators supported away from the two balls used to form the connected sum, so mixed cup products vanish while each square gives the common orientation class. WriteIf $\mathrm{deg}h=n$, the three cup product relations sayTaking determinants gives $-(\det P{)}^2=n^2$, which forces $n=0$. Therefore the only possible degree to the connected sum is $0$, realized by a constant map. This is a degree constraint from intersection forms: the indefinite intersection form of the sphere product cannot pull back a definite form with a nonzero degree multiplier.

The $q$-skeleton of a CW complex is the subcomplex consisting of its topological cells of dimension at most $q$. A cellular map preserves these skeleta. The cellular approximation theorem lets continuous maps respect this filtration up to homotopy.