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For a continuous map $f:X\to Y$, precomposition with the induced map on singular chains sends a singular cochain on $Y$ to one on $X$. It commutes with the coboundary and induces the induced map on cohomology.

For any continuous map between the displayed Real projective spaces, the induced map on cohomology sends the degree-one generator to $\alpha x+\beta y$. Its $2n$th power is $\left(\genfrac{}{}{0ex}{}{2n}{n}\right){\alpha }^n{\beta }^n{x}^n{y}^n$, since all other terms exceed the truncation exponent of one factor. The coefficient is zero modulo two because $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=2\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right)$. Naturality of the evaluation pairing then makes the top induced map on homology zero. This obstruction does not apply to $n=0$, when both spaces are points.

Let $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$ be the standard inclusion. If $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ and $a=w_1({\gamma }_d)$, thenThe restricted complex tautological bundle is the complexification of a real vector bundle ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. Its underlying real bundle is ${\gamma }_d\oplus {\gamma }_d$. The Whitney product formula for Stiefel–Whitney classes gives total class $(1+a{)}^2=1+a^2$, while the Stiefel–Whitney class of the underlying real bundle of a complex line identifies its second class with the reduced First Chern class. Dualizing changes the integral sign but has no effect modulo two. Thus the full induced map on cohomology sends $h^j$ to $a^{2j}$.

Work with coefficients ${\mathbb{F}}_2$. For $n\ge 1$, the mod-two cohomology ring of real projective space and the Künneth theorem givewhere all three generators have degree one. Any induced map on cohomology has $f^{\ast }z=\alpha x+\beta y$, with $\alpha ,\beta \in {\mathbb{F}}_2$. By naturality of the cup product, the only possible term in top degree isAll other terms vanish because their exponent of $x$ or $y$ exceeds $n$. Butis even for every $n\ge 1$. Thus the displayed top-degree induced map on cohomology is zero for every $f$.

Both top-degree homology groups are one-dimensional over ${\mathbb{F}}_2$, and their dual cohomology groups pair with them nondegenerately. Naturality of this evaluation makes $f_{\ast }$ dual to $f^{\ast }$, so $f_{\ast }$ is also zero and cannot be an isomorphism. This is the mod-two degree obstruction for equal projective factors. ConsequentlyIf zero is included among the allowed values, ${\mathbb{RP}}^0$ is a point and the unique map is an isomorphism on $H_0$. In that convention, the only value is $n=0$.

Use the standard orientations on the three spheres and the complex orientation on Complex projective space. Let $h\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ satisfy $⟨h^3,[{\mathbb{CP}}^3]⟩=1$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^4)$. The Künneth theorem givesWrite the induced map on cohomology as $f^{\ast }h=pa+qb+rc$ with integers $p,q,r$. Naturality and graded commutativity of the cup product giveThe mapping degree is therefore $d=6pqr$. This is the factorial degree obstruction for products of two-spheres. Every positive value is a positive multiple of six, so the least possible positive degree is at least $6$.

To attain it, identify each oriented sphere with ${\mathbb{CP}}^1$. Send three projective linear forms to the coefficients of their product:whereNo product of nonzero linear forms is the zero polynomial. Rescaling any factor rescales all coefficients by the same nonzero scalar. The map is therefore well-defined and is a holomorphic map on the products of projective charts.

Choose a homogeneous cubic with three distinct roots, all in an affine chart. Its preimages are exactly the $3!=6$ orderings of its three projective linear factors. In root coordinates, the derivative of the elementary-symmetric coefficient map has determinant equal, up to sign, to the Vandermonde determinant, the product of the three pairwise root differences. This is nonzero for distinct roots, so all six preimages are regular. A nonsingular complex-linear derivative preserves the underlying real orientation, since its real determinant is the squared absolute value of its complex determinant. Each preimage consequently contributes $+1$ to the mapping degree. HenceThis polynomial multiplication map to complex projective space provides the requested example without assuming that a prescribed cohomology homomorphism is realizable by a continuous map.

Let $a$ and $b$ be the pullbacks of the standard generator of $H^n(S^n;\mathbb{Z})$ from the two factors. The Künneth theorem and graded commutativity of the cup product give

Each homeomorphism acts invertibly on $H^n\cong {\mathbb{Z}}^2$. To respect ordinary composition, send $h$ to $(h^{-1}{)}^{\ast }$; functoriality of induced map on cohomology then defines a homomorphism

For $n=1$, the space is the torus. Every matrix in $GL(2,\mathbb{Z})$ induces a linear homeomorphism ${\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{R}}^2/{\mathbb{Z}}^2$, so the image is all of $GL(2,\mathbb{Z})$.

For $n=2$, write $h^{\ast }a=pa+qb$. Since $a^2=0$ and $ab=ba$,so $pq=0$; the same argument applies to $h^{\ast }b$. Invertibility then forces the matrix to be a signed permutation matrix. Every such matrix is realized by swapping the two sphere factors and applying an orientation-reversing homeomorphism to either factor. Thus the image consists exactly of the eight signed permutation matrices.