Let $X$ be a topological space, and $A \subseteq X$ a subspace. How to think about an element $u \in H^n(X, A)$? Is the following correct? $u$ can be represented by a function $U$ taking an $n$-cell $\sigma$ in $X$ as input and assigning an integer $U(\sigma)$ as output. If $\sigma \subseteq A$, then $U(\sigma)=0$. Also, $U$ applied to the boundary $\partial \tau$ of any $n+1$-cell $\tau$ of $X$ is $0$. Finally, we can replace $U$ by $U+\delta V$, which assigns to $\sigma$ the value $U(\sigma)$ plus $V(\partial \sigma)$, where $V$ is a function on $n-1$ cells that vanishes on cells in $A$.

So the value of the ${\textit class}$ $u$ on $\sigma$ is not well defined.

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    $\begingroup$ I want to give an answer in terms of Brown representability, are you familiar with it? $\endgroup$ – Benjamin Gadoua Jun 30 '16 at 1:11
  • $\begingroup$ Vaguely familiar - under certain conditions, we get a cohomology theory represented by a space/spectrum, and vice versa $\endgroup$ – usr0192 Jun 30 '16 at 1:12
  • $\begingroup$ Yeah, we're only going to use basic ideas here since the question is in terms of ordinary homology, so we just need the very concrete idea that $H^n(-;\mathbb{Z})$ is represented by the spectra of spheres. $\endgroup$ – Benjamin Gadoua Jun 30 '16 at 1:15
  • $\begingroup$ Do you know about manifolds and duality theorems? $\endgroup$ – Daniel Valenzuela Jun 30 '16 at 13:43
  • $\begingroup$ @DanielValenzuela not really. I mean, I've read them a long time ago and tried reading the proof of poincare duality, but it seemed complicated $\endgroup$ – usr0192 Jun 30 '16 at 13:45

Classically Brown Representability says that if we have a cohomology theory $k^*$ satisfying the Meyer-Vietoris axiom and the wedge axiom that we can find a classifying spectrum $Y_n$ and element $u_n$ of $k^n(Y_n)$ where $B_{u_n}:H^n(-) \to [-:Y_n,y_*]$ is a natural equivalence. This is the case for reduced cohomology. To make this applicable to your comment, we want to look at ordinary cohomology, which we have the Eilenberg-Maclane spaces for, a notational note, I will use $H(n,G)$ instead of the usual $K$.

Recall that there is a natural equivalence $T^n:H^n \to \tilde{H^n}$ for ordinary homology that takes $H^n(X,A) \mapsto \tilde{H^n}(X/A)$. Combining these two ideas we get that $H^n(X,A;G) \cong [X/A,*:H(n,G)]$.

Of course, as one commentator mentions, we can find other ways to get cohomology classes by looking at duality and orientation theorems, or by looking at other cohomology theories and natural transformations to ordinary cohomology.

  • $\begingroup$ Ok, but is the way I suggested of representing cohomology classes incorrect? $\endgroup$ – usr0192 Jun 30 '16 at 21:25
  • $\begingroup$ @user90219, your suggested representation is in fact the definition of relative (singular/simplicial) cohomology. $\endgroup$ – Mariano Suárez-Álvarez Jun 30 '16 at 21:27
  • $\begingroup$ Yes, that's what I was unwinding to form my answer. Sorry for the confusion! $\endgroup$ – usr0192 Jun 30 '16 at 21:29
  • $\begingroup$ I think we are fundamentally describing the idea, vanishing on $A$ would describe the pointed homotopies with $A$ quotientied to a point. as Mariano says, this is directly because of the definition. Usually when one is asking about representation of (co)homology classes you are looking for some way to turn them into maps or objects you can get your hands on with some other type of theory. $\endgroup$ – Benjamin Gadoua Jun 30 '16 at 21:29

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