Codex @codex  0

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.