----- > [!proposition] Proposition. ([[singular cohomology is not dualized singular homology]]) > There are natural maps $H^{n}(X) \to \text{Hom}\big( H_{n}(X), \mathbb{Z} \big)$, and similarly for [[relative singular homology]], which commute with the $\partial{}$-maps in the [[long exact sequence for relative singular homology|long exact sequence for a pair]]. These maps are always [[surjection|surjective]], but are not always [[isomorphism|isomorphisms]]. ^proposition > [!proof]- Proof. ([[singular cohomology is not dualized singular homology]]) > Define $\begin{align} > H^{n}(X) &\to \text{Hom}\big( H_{n}(X), \mathbb{Z} \big) \\ > [\alpha]& \mapsto \big([a] \mapsto \alpha(a) \big) > \end{align}$ > this is [[well-defined]] because $\begin{align} > [a]=[b] &\implies a-b = dc \text{ for some }c \\ > &\implies \alpha(a-b) = \alpha(d c) \\ > & \implies \alpha(a) - \alpha(b) = (d\alpha)(c) = 0 \\ > & \implies \alpha(a)=\alpha(b) > \end{align}$ > where we have used that $\alpha$ is a cocycle (so $d\alpha=0$) and that $d$ commutes with it, in the sense that in general (and by definition) $(d^{i}\alpha)(c)=d_{i}^{*}(\alpha)=\alpha \circ d_{i}(c)$. And to show invariance to choice of representative, suppose $[\alpha]=[\beta]$; then $\alpha - \beta=d \gamma$ for some $\gamma$, whence $\begin{align} > [\alpha - \beta] &\mapsto \big([a] \mapsto (d \gamma)(a)\big) \text{, that is,} > \\ [\alpha - \beta] & \mapsto \big( [a] \mapsto 0 \big) > \end{align}$ > which means $[\alpha - \beta]=0$. > > To see surjectivity, let $\psi \in \text{Hom}\big( H_{n}(X) , \mathbb{Z} \big)$. Define $\begin{align} > \alpha: C_{n}(X) &\to \mathbb{Z} \\ > a &\mapsto \psi([a]) > \end{align}$ > and note that, under the given map, $[\alpha]$ maps to $\big( [a] \mapsto \alpha(a)=\psi([a]) \big)$, i.e., $[\alpha]$ is in the preimage of $\psi$. > > The map is not an isomorphism, however: consider the [[chain complex of modules|chain complex]] $0 \to \mathbb{Z} \xrightarrow{0}\mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0$ > which dualizes to $0 \xleftarrow{} \mathbb{Z} \xleftarrow{0} \mathbb{Z} \xleftarrow{2} \mathbb{Z} \xleftarrow{0} \mathbb{Z} \xleftarrow{}\textcolor{Skyblue}{0}.$ So its cohomology is $\textcolor{Skyblue}{0},\mathbb{Z}, 0, \mathbb{Z}/2, \mathbb{Z}, 0$. If an isomorphism with the desired property existed, then we'd have $0\cong \text{Hom}(\mathbb{Z}, \mathbb{Z})$. ----- #### ---- #### 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 > ```