---- > [!definition] Definition. ([[singular cohomology]]) > Let $X$ be a [[topological space]]. [[dual vector space|Dualizing]] [[dual map|the]] [[singular homology|singular]] [[chain complex of modules|chain complex]] $(C_{\bullet}, d_{\bullet})$ on $X$ yields a [[chain complex of modules|cochain complex]] $(C^{\bullet}, d^{\bullet})$ $C^{0}(X) \xrightarrow{d^{0}} C^{1}(X) \xrightarrow{d^{1}} \cdots C^{k}(X) \xrightarrow{d^{k}} C^{k+1}(X) \to \cdots$ whose [[(co)homology of a complex|cohomology]] $H^{i}(X):=H^{i}(C^{\bullet}, d^{\bullet})$ is called the **singular cohomology** of the space $X$. > > > Explicitly, we have cochain groups $C^{k}(X)=\text{Hom}_{\mathbb{Z}\text{-}\mathsf{Mod}}\big(C_{k}(X), \mathbb{Z}\big)$ > and codifferential $\begin{align} > d^{k}(X): \text{Hom}\big(C_{k}(X), \mathbb{Z} \big)& \to \text{Hom}\big( C_{k+1}(X), \mathbb{Z} \big) \\ > \big(C_{k}(X) \xrightarrow{\varphi} \mathbb{Z} \big)& \mapsto \big(C_{k+1}(X) \xrightarrow{d_{k+1}} C_{k}(X) \xrightarrow{\varphi} \mathbb{Z}\big ). > \end{align}$ > > > $H^{n}(-)$ is a [[contravariant functor|contravariant]] [[covariant functor|functor]] $\mathsf{Top} \to \mathsf{Ab}$ (or $\mathsf{Top} \to R\text{-}\mathsf{Mod}$ if considering [[(co)homology with coefficients|homology with coefficients]] in a general [[commutative ring|commutative]] [[ring]] $R$), obtained by composing the (contravariant!) '[[singular (co)chain map and homomorphism induced by a continuous map|singular cochain group functor]]' $C^{n}(-):\mathsf{Top} \to \mathsf{Cochain}(\mathsf{Ab})$ with the (covariant) $n$th [[homomorphism on cohomology induced by a cochain map|cohomology functor]] $\mathsf{Cochain}(\mathsf{Ab}) \to \mathsf{Ab}$. > > [!definition] Definition. (The singular cohomology ring) > The singular cohomology $H^{*}(X; R) := \bigoplus_{n \geq 0}H^{n}(X; R)$of $X$ with coefficients in a [[ring]] $R$ has the structure of a [[graded ring]] (and [[graded module|graded]] $R$-[[module]]), where multiplication is given by the [[cup product]] $- \smile -: H^{k}(X ;R) \times H^{\ell}(X; R) \to H^{k+\ell}(X ;R).$ Indeed, the map $\smile$ is [[associative]] on [[chain complex of modules|cochains]], hence on $H^{*}$. There is also a map $1:C_{0}(X) \to R$ sending $\sigma \mapsto 1_{R}$ for [[singular simplex|all]] $\sigma:\Delta^{0} \to X$. Then we have $[1] \smile [\varphi]=[ 1 \smile \varphi]=[\varphi].$ Thus there is unital ring structure indeed. *This ring is not commutative!* Instead one has the following "graded commutativity": if $\alpha \in H^{k}(X; R)$ and $\beta \in H^{\ell}(X; R)$, then $\alpha \smile \beta=(-1)^{k \ell} \beta \smile \alpha.$ Note that this "graded commutativity" is only true for cohomology classes — it is not true in general for cochains. ^definition > [!note] Remark. > Singular cohomology $H^{\bullet}(X)$ is *not* [[singular cohomology is not dualized singular homology| just dualized singular homology]]. In fact, $H^{\bullet}(X)$ is better than $H_{\bullet}(X)$ because the former carries the structure of a [[graded ring]], via the [[cup product]] construction. The two interact e.g. in the context of [[Poincare duality]]. > > > > [!basicexample] Example. ($\mathbb{C}P^{2} \text{ vs. }\mathbb{S}^{2} \vee \mathbb{S}^{4}$) > > The singular (co)homology groups [[complex projective space|for]] $\mathbb{C}P^{2}$ equal those for $\mathbb{S}^{2} \vee \mathbb{S}^{4}$, [[reduced homology of a good sum of wedges is direct sum of reduced homologies|namely]], $H^{i}(\mathbb{C}P^{2}; \mathbb{Z}) = \begin{cases} > > \mathbb{Z} \langle 1 \rangle & i=0 \\ > > \mathbb{Z} \langle a \rangle & i =2 \\ > > \mathbb{Z}\langle b \rangle & i =4 \\ > > 0 & \text{else}. > > \end{cases}$ > > So (co)homology groups alone cannot help us distinguish between the two spaces. Investigating the cohomology *[[ring]]* structure is what allows us to deduce that the spaces are not [[homotopy equivalent]]. > > > > Indeed, [[Gysin sequence|we know]] that $a \smile a=b$ in $H^{*}(\mathbb{C}P^{2}; \mathbb{Z})$, thus for any map $f:\mathbb{S}^{4} \to \mathbb{C}P^{2}$, can [[homomorphism on cohomology induced by a cochain map|pull back]] the class $b \in H^{4}(\mathbb{C}P^{2})$ as $f^{*}(b)=f^{*}(a \smile a)=f^{*}(a) \smile f^{*}(a),$ > > where we have [[cup product|used that]] $f^{*}:H^{*}(\mathbb{C}P^{2}; \mathbb{Z}) \to H^{*}(\mathbb{S}^{4}; \mathbb{Z})$ is a [[ring homomorphism]]. Of course, $f^{*}(a) \in H^{2}(\mathbb{S}^{4}; \mathbb{Z})=0$ is trivial, hence $f^{*}(b)$ is trivial. Since $b$ [[submodule generated by a subset|generates]], it follows that any map $f:\mathbb{S}^{4} \to \mathbb{C}P^{2}$ induces the zero map on $4$th cohomology. > > > > So $\mathbb{C}P^{2} \not \simeq \mathbb{S}^{2} \vee \mathbb{S}^{4}$: e.g. because the inclusion $\mathbb{S}^{4} \hookrightarrow \mathbb{S}^{2} \vee \mathbb{S}^{4}$ doesn't induce zero map on $4$th cohomology. > > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```