Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 20H (Algebraic Topology)

Let $X=Y\cup Z$, where $Y$, $Z$, and $Y\cap Z$ are path-connected subcomplexes containing the basepoint; small open neighbourhoods may be used to reduce to the usual open-cover statement. The Seifert-van Kampen theorem says thatMore explicitly, this is the quotient of the free product ${\pi }_1(Y)\ast {\pi }_1(Z)$ by the normal closure ofEquivalently, ${\pi }_1(X)$ is the pushout of the two inclusion-induced homomorphisms.

The wedge intersection is simply connected after taking suitable neighbourhoods, so repeated application of the Seifert-van Kampen theorem gives the free product of $n$ copies of $\mathbb{Z}$ withThus, for any basepoint,This is the fundamental group of a wedge of circles and a real projective plane.

Let $V_4=⟨u,v\mid u^2=v^2=[u,v]=1⟩$. Define a surjectionIts kernel determines a connected regular four-sheeted cover of a wedge of circles and a real projective plane, say ${\tilde{X}}_n\to X_n$.

Use the CW structure on $X_n$ with one vertex, one edge for every $a_i$ and for $b$, and one two-cell attached along $b^2$. In the regular cover, the degree-two cellular chain group isThe cellular boundary formula gives the lifted attaching word $b^2$ the boundaryin the $b$-edge summand of $C_1$. ThereforeThere are no three-cells, so

If $X_n$ were homotopy equivalent to a compact connected surface $S$, the subgroup corresponding to this cover would produce a connected four-sheeted cover $\tilde{S}$ homotopy equivalent to ${\tilde{X}}_n$. A finite cover of a compact surface is again a compact connected surface. But the integral top homology of a compact connected surface has rank at most one, contradicting the displayed rank two. Hence $X_n$ is not homotopy equivalent to a compact connected surface for any $n\ge 1$.