Algebraic Topology II - 2020
Fabian, 17.08.2020, 11:00-11:30
https://drive.google.com/file/d/1NzTIdKo5q6vGP7gc8Q1PV7SWbEcaJvwy/view?usp=sharing
Sascha, 24.08.2020, 11:00–11:30
Half an hour before the exam I was led into a separate room and given the following question:
Let \(M\) be a closed connected oriented \(n\)-manifold and suppose that there exists a map \(f\colon S^n \to M\) of non-zero degree. Show that \(H_*(M;\mathbb{Q}) \cong H_*(S^n;\mathbb{Q})\). If the degree is \(\pm 1\), show that \(H_*(M;\mathbb{Z}) \cong H_*(S^n;\mathbb{Z})\). Hint: Poincaré duality might help.
I was then given 30 minutes to prepare a solution to this exercise, using any notes I brought with me. I prepared the following solution:
Both \(M, S^n\) are n-dimensional, closed, connected and oriented, so the 0-th and n-th homology groups with \(R\)-coefficients are isomorphic to \(R\) for both. Let now \(0 < i < n\) and let \(k := \mathrm{deg}(f)\). By Poincaré-duality, the map \(H^i(M;R) \to H_{n-i}(M;R)\) given by \([\alpha] \mapsto [M] \frown [\alpha]\) is an isomorphism for all \(i\). Let \([\alpha] \in H^i(M;R)\). Then we compute:
$$ [M] \frown k[\alpha] = k[M] \frown [\alpha] = f_*[S^n] \frown [\alpha] = f_*([S^n] \frown f^*[\alpha]) = 0,$$
using bilinearity of \(\frown\), definition of degree, naturality of the cap product and finally that \(H_{n-i}(S^n;R) = 0\).
From this we conclude that \(k[\alpha] = 0\). If \(R = \mathbb{Q}\) or \(R = \mathbb{Z}\) and \(k = \pm 1\), this implies \([\alpha] = 0\), so the \(i\)-th cohomology and therefore the \((n-i)\)-th homology group of \(M\) is trivial, same as that of \(S^n\).
At 11:00, I was led into the other room, where I presented this solution to Prof. Sisto on the blackboard. He let me explain without interruption and commented that it was “perfect”. He then pondered what would happen if we replaced \(S^n\) with a more general manifold \(N\) (that is still closed, connected and oriented). He said he hadn’t thought about this before but thought that there might be an inequality for the ranks of the homology groups. We then worked through that problem together, with him giving a few hints, until we were able to conclude that \(f_*\) must be surjective, from which it then follows that \(\mathrm{rank}H_i(N) \geq \mathrm{rank}H_i(M)\). He then said we were done, asked me what plans I had for my future studies and let me go about fifteen minutes early.
