Cohomology Rings or: How I Learned to Stop Worrying and Love the Cup Product

Let's talk about fun ways we can translate topological problems into algebraic ones. Suppose we have an $(n-1)$-sphere, i.e. $S^{n-1}$ and the $n$-disk $D^n$. It's well known that these two spaces are very different from each other, but can we retract one of these to the other? One way we can approach this is by applying a homology functor. Since $S^{n-1}$ is a subspace of $D^n$ such that $\partial D^n = S^{n-1}$, i.e. $S^{n-1}$ is the boundary of the $n$-disk, suppose there were a retraction $r: D^n \to S^{n-1}$. Can we do this? Recall that a retraction is such that we get the following mapping:

where $i: A \to X$ is the inclusion map and $r: X \to A$ is a retraction such that $r \circ i = \text{id}_{A}$. Note that this follows from the property of retractions where $r(a) = a$ for all $a \in A$. Going back to $S^{n-1}$, we have a similar mapping:

where $i: S^{n-1} \to D^n$ just comes from the boundary inclusion $S^{n-1} \subset D^n$. Next, we apply the homology functor:

Note that we're using $\mathbb{Z}$ coefficients for convenience. Let's focus on the $(n-1)$ homology group, $H_{n-1}$:

Since $D^n$ is contractible, we know that $H_{n-1}(D^n;\mathbb{Z}) \cong 0$ for $n > 1$. Similarly, we know that $H_{n-1}(S^{n-1}; \mathbb{Z}) \cong \mathbb{Z}$, hence we get:

where $i_*$ and $r_*$ are the induced maps on homology for the inclusion and retraction maps, respectively. Recall that $r \circ i = \text{id}_{S^{n-1}}$, so applying the homology functor to this, we get

where $\text{id}_{H_*(S^{n-1})}$ is just the identity map on the homology of $S^{n-1}$. However, we see that for $\mathbb{Z} \xrightarrow{\,\, i_* \,\,} \{ 0 \} \xrightarrow{\,\, r_* \,\,} \mathbb{Z}$, this mapping factors through zero which means the overall map must be the zero map, i.e. the map $\mathbb{Z} \to \{ 0 \}$ must map every integer to zero, and then likewise $\{ 0 \} \to \mathbb{Z}$ also just maps $0 \mapsto 0$. This is the zero homomorphism. Note that $\text{id}_{H_*(S^{n-1})}$ is not the zero homomorphism since the zero homomorphism is not the identity map of $H_*(S^{n-1})$ (suppose we have some non-zero $a \in H_{n-1}(S^{n-1})$, then $0(a) \neq a$ where $0(-)$ denotes the zero homomorphism), so we've reached a contradiction. Since we initially assumed that this retraction $r: D^n \to S^{n-1}$ existed, we see that this must not be the case. Thus, a retraction from the $n$-disk to the $(n-1)$-sphere does not exist.

The example above showcases how we can use homology to convert this retraction problem into one about induced function composition. As we'll see, homology is not all-powerful, and we'll need to develop a bit more advanced machinery to tackle more interesting problems.

Let's see if homology can help us with the following problem: Does there exist a continuous map $f: \mathbb{RP}^m \to \mathbb{RP}^n$ for $1 < n < m$ such that it induces an isomorphism on fundamental groups, i.e.

where $\mathbb{RP}^k$ is $k$-dimensional real projective space? Recall that real projective space is such that $\pi_1(\mathbb{RP}^k) \cong \mathbb{Z}_2$ for all $k >1$ and since $\mathbb{RP}^1$ is homeomorphic to $S^1$, then we have $\pi_1(\mathbb{RP}^1) \cong \pi_1(S^1) \cong \mathbb{Z}$ for the $k=1$ case. On the homology side of things, we have $H_0(\mathbb{RP}^k) \cong \mathbb{Z}$ and $H_{k}(\mathbb{RP}^k) \cong \mathbb{Z}$ if $k$ is odd (note that $\mathbb{RP}^k$ for odd $k$ is orientable, but $\mathbb{RP}^k$ for even $k$ is non-orientable). Then we also have $H_{i}(\mathbb{RP}^k) \cong \mathbb{Z}_2$ if $0 < i < k$ and $i$ is odd and $H_{i}(\mathbb{RP}^k) \cong 0$ if $i$ is even. Note that $H_k(\mathbb{RP}^{k}) \cong 0$ for $k > 0$ when $k$ is even (note: here we're emphasizing that the homology group is zero even at the top-most dimension). From the fundamental group perspective, there's no clear obstructions since we just have the induced map:

so the fundamental group doesn't distinguish between $\mathbb{RP}^n$ versus $\mathbb{RP}^m$ for $n \neq m$ and $1 < n < m$. Note that if $n=1$ then this problem becomes a lot easier since $\mathbb{Z} \neq \mathbb{Z}_2$.

What about for homology? We have:

Looking at just the $H_1$ case we have an induced map of $f_*: H_1(\mathbb{RP}^m) \to H_1(\mathbb{RP}^n)$ such that we get the same induced map from before with $f_*: \mathbb{Z}_2 \to \mathbb{Z}_2.$ Can homology distinguish between these spaces more than the fundamental group? Well, suppose $m$ were odd and $n$ were even, then for the $m$-th homology group we'd get:

where $H_m(\mathbb{RP}^m) \cong \mathbb{Z}$ and $H_m(\mathbb{RP}^n) \cong 0$ since $m > n$, so we just get the zero map $f_*: \mathbb{Z} \to \{0\}$. Note that this does not imply that $f$ can't exist. $f_*: \mathbb{Z} \to \{0\}$ is only a constraint on $f$. Additionally, note that even if we consider a mix of these odd/even cases for $n$ and $m$, these would merely be placing constraints on $f$ and not leading to any kind of contradiction we could use to conclude that $f$ can't possibly exist. Here's where we run into a dead end with homology.

What about cohomology? Before we jump into $H^*(\mathbb{RP}^k;-)$, let's build some intuition by first looking at $H^*(S^1;\mathbb{Z})$. Recall that $H^0(S^1;\mathbb{Z}) \cong \mathbb{Z}\langle 1\rangle$, where we have a generator $1 \in H^0(S^1;\mathbb{Z})$. Next, we have $H^1(S^1;\mathbb{Z}) \cong \mathbb{Z}\langle\alpha\rangle$, where we have a different generator $\alpha \in H^1(S^1;\mathbb{Z})$. Note that $H^k(S^1;\mathbb{Z}) \cong 0$ for all $k > 1$. With this, we see that the total cohomology $H^*(S^1;\mathbb{Z})$ is generated as a group by $1$ and $\alpha$. Furthermore, note that the generator $1 \in H^0(S^1;\mathbb{Z})$ acts as the multiplicative identity for the entire cohomology ring, i.e.

where we have the standard cohomology cup product $\smile: H^n(X;-) \times H^m(X;-) \to H^{n+m}(X;-)$. Importantly, since $\alpha^2 = \alpha \smile \alpha \in H^2(S^1;\mathbb{Z})$ and $H^2(S^1;\mathbb{Z}) \cong 0$, then we get that $\alpha^2 = 0$. This is the central relation we want to capture for the cohomology ring $H^*(S^1;\mathbb{Z})$. We can get this with the truncated polynomial ring $\mathbb{Z}[\alpha] / (\alpha^2)$ where we mod out by the ideal $(\alpha^2) \subset \mathbb{Z}[\alpha]$ to capture the relation $\alpha^2 = 0$. Putting this all together, we get

where elements $b_1 + b_2\alpha, c_1+c_2\alpha \in H^*(S^1;\mathbb{Z})$ are such that

Note that this is the simplest non-trivial example of an exterior algebra which is denoted as either $\Lambda_\mathbb{Z}[\alpha]$ or just $\Lambda[\alpha]$ if the ring that's being worked over is obvious.

Next, let's consider $H^*(\mathbb{RP}^2;\mathbb{Z}_2)$ with $\mathbb{Z}_2$ coefficients. Recall that $H^0(\mathbb{RP}^2;\mathbb{Z}_2) \cong \mathbb{Z}_2\langle 1\rangle$, $H^1(\mathbb{RP}^2;\mathbb{Z}_2) \cong \mathbb{Z}_2\langle\alpha\rangle$, $H^2(\mathbb{RP}^2;\mathbb{Z}_2) \cong \mathbb{Z}_2\langle\beta\rangle$, and $H^k(\mathbb{RP}^2;\mathbb{Z}_2) \cong 0$ for all $k > 2$, where we have non-zero generators $\alpha \in H^1(\mathbb{RP}^2;\mathbb{Z}_2)$ and $\beta \in H^2(\mathbb{RP}^2;\mathbb{Z}_2)$. Note that we have $\alpha \smile \alpha = \beta$ (non-zero generators mapping to non-zero generators), and since $H^3(\mathbb{RP}^2;\mathbb{Z}_2) = 0$ then $\alpha^3 = \alpha \smile (\alpha \smile \alpha) = 0$. Similar to the $H^*(S^1;\mathbb{Z})$ case, we want to capture this $\alpha^3 = 0$ relation. This gives us the following truncated polynomial ring:

From this, we see how this generalizes to the $\mathbb{RP}^n$ case, where $H^k(\mathbb{RP}^n;\mathbb{Z}_2) \cong 0$ for all $k > n$, so we want to capture the relation $\alpha^{n+1} = 0$ where $\alpha \in H^1(\mathbb{RP}^n;\mathbb{Z}_2)$ is the non-zero generator, thus we get the following cohomology ring:

Taking a small step back, if we start with $f: \mathbb{RP}^m \to \mathbb{RP}^n$ then we can consider the induced map on cohomology rings

note that we're mapping $H^*(\mathbb{RP}^n;\mathbb{Z}_2) \mapsto H^*(\mathbb{RP}^m; \mathbb{Z}_2)$ since cohomology is a contravariant functor. Let $\alpha_n \in H^1(\mathbb{RP}^n;\mathbb{Z}_2)$ be the generator for this cohomology group. How would $f^*$ map $\alpha_n$? One way we can see this is by considering the Universal Coefficient Theorem, where we have an isomorphism

so in the $H^1(\mathbb{RP}^n;\mathbb{Z}_2)$ case we have: $H^1(\mathbb{RP}^n;\mathbb{Z}_2) \cong \text{Hom}(H_1(\mathbb{RP}^n;\mathbb{Z}_2), \mathbb{Z}_2)$ since $\mathbb{Z}_2$ is a field. Note that this does not hold for $\mathbb{Z}.$ With this, we get the following commutative diagram

where $f^*$ is the previous induced mapping on cohomology and $(f_*)^*$ is the induced map we get after applying the $\text{Hom}$ functor to the induced map on homology $f_*: H_1(\mathbb{RP}^m;\mathbb{Z}_2) \to H_1(\mathbb{RP}^n;\mathbb{Z}_2)$, where we get

since $\text{Hom}$ is also a contravariant functor similar to cohomology. Let's trace where the generator $\alpha_n \in H^1(\mathbb{RP}^n;\mathbb{Z}_2)$ gets mapped.

Let $\alpha_n \in H^1(\mathbb{RP}^n;\mathbb{Z}_2)$. $\alpha_n \mapsto f^*(\alpha_n) \in H^1(\mathbb{RP}^m;\mathbb{Z}_2)$. $f^*(\alpha_n) \mapsto \Psi_m( f^*(\alpha_n) ) \in \text{Hom}(H_1(\mathbb{RP}^m;\mathbb{Z}_2), \mathbb{Z}_2)$. Here, we will denote $\Psi_m( f^*(\alpha_n) )$ as the non-zero homomorphism $\phi_m$ such that

i.e. $\Psi_m( f^*(\alpha_n) ) = \phi_m$. Next, going down then right, we get: $\alpha_n \mapsto \Psi_n(\alpha_n)$, which is just the corresponding non-zero homomorphism $\phi_n: H_1(\mathbb{RP}^n, \mathbb{Z}_2) \to \mathbb{Z}_2$, so we have $\alpha_n \mapsto \Psi_n(\alpha_n) = \phi_n$. Then, $\phi_n \mapsto (f_*)^*(\phi_n)$, where we pullback $\phi_n$ (since contravariant) to get:

and similarly to $\Psi_m( f^*(\alpha_n) )$, we must have that $\phi_n \circ f_*$ is also the non-zero homomorphism $\phi_m: H_1(\mathbb{RP}^m, \mathbb{Z}_2) \to \mathbb{Z}_2$ where $\phi_m \in \text{Hom}(H_1(\mathbb{RP}^m;\mathbb{Z}_2), \mathbb{Z}_2)$. Putting things together, we get:

and since $\phi_m = \Psi_m(\alpha_m)$, then $\Psi_m(f^*(\alpha_n)) = \phi_m$ implies that $\Psi_m(f^*(\alpha_n)) = \Psi_m(\alpha_m)$, which means that

With this, we see that $f^*$ maps the generator $\alpha_n \in H^1(\mathbb{RP}^n;\mathbb{Z}_2)$ to the generator $\alpha_m \in H^1(\mathbb{RP}^m;\mathbb{Z}_2)$. Why is this an issue? Well, we have two different cohomology rings with different relations. In $H^*(\mathbb{RP}^n;\mathbb{Z}_2)$ we have that $\alpha_n^{n+1} = 0$ and in $H^*(\mathbb{RP}^m;\mathbb{Z}_2)$ we have that $\alpha_m^{m+1} = 0.$ Since we have $1 < n < m,$ then consider

Note that $\alpha_m^m \in H^m(\mathbb{RP}^m;\mathbb{Z}_2)$ is the non-zero generator for this cohomology group, but if we apply $f^*$ to $\alpha_n^{m}$ then we get:

where $\alpha_m^m$ is non-zero, but recall that $\alpha_n^m = 0$, so this contradicts $f^*(\alpha_n^m) = f^*(0) = 0$.

TL;DR: In $H^*(\mathbb{RP}^n;\mathbb{Z}_2)$, the element $\alpha_n^m = 0$ since $m > n$ and $H^m(\mathbb{RP}^n;\mathbb{Z}_2)$ is trivial. However, when we map this "zero" to $H^*(\mathbb{RP}^m;\mathbb{Z}_2)$ via $f^*$ we get $f^*(\alpha_n^m) = (f^*(\alpha_n))^m = \alpha_m^m$, where $\alpha_m^m \in H^m(\mathbb{RP}^m;\mathbb{Z}_2)$ is a non-zero generator in $H^*(\mathbb{RP}^m;\mathbb{Z}_2)$ which contradicts $f^*(\alpha_n^m) = f^*(0) = 0$ being zero.

With this, we see through the cohomology rings of $\mathbb{RP}^n$ and $\mathbb{RP}^m$ that the function $f: \mathbb{RP}^m \to \mathbb{RP}^n$ cannot exist such that it induces an isomorphism $f_*: \pi_1(\mathbb{RP}^m) \to \pi_1(\mathbb{RP}^n)$. ■