**Theorem:**

Let $K$ be a complex with $r$ combinatorial components. Then $H_{0}(K)$ is isomorphic to the direct sum of (not $r$ but) $2r$ copies of the group $Z$ of integers.

**Proof for one component:**

As 0-cycles are synonymous with 0-chains, they have $\alpha_{0}$ many dimensions-of-freedom for sure. How many the 0-boundaries of our component have? Since the incidence matrix is a $\alpha_{0}$ by $\alpha_{1}$ matrix (see my previous posting with ‘homology’ tag), they have at least $\alpha_{1} - \alpha_{0}$ many for sure. And 1 more for the rank deficiency! The proof follows from this arithmetic.

]]>If one solves

${b}_{1}\u27e8{a}_{1}\u27e9+\dots +{b}_{m}\u27e8{a}_{m}\u27e9=\delta [\sum \sum {x}_{i,j}\u27e8{a}_{i},{a}_{j}\u27e9]$for x’s when b’s are given, one ends up with

$\mathit{\eta}\overrightarrow{x}=\overrightarrow{b}$where \eta is the INCIDENCE MATRIX! Then all proofs about 0-homology groups are just about the ranks of square-submatrices of this matrix. Am I right?

]]>In reduced homology#relation to relative homology, at the bottom is a computation where one step is that $\ker H_0(\epsilon) \cong \operatorname{coker}{H_0(x)}.$ Can you explain that step to me? If you think of kernels and cokernels as morphisms, then there’s no way such an isomorphism can hold, since $\ker H_0(\epsilon)$ is a morphism into $H_0(X)$, while $\operatorname{coker}{H_0(x)}$ is a morphism *out of* $H_0(X).$ But if you think of kernels and cokernels as objects, then I guess it’s possible for them to be isomorphic. My guess is it must follow from the fact that $H_0(\epsilon)\circ H_0(x)$ is an iso, so the statement is something like “the kernel of a split epimorphism is the cokernel of its right-inverse,” but I can’t figure it out.

If generalized nonabelian cohomology, from the nPOV, means a hom-space in some $(\infty,1)$-topos, then it can equivalently be characterized as global sections of an object in some $(\infty,1)$-topos, since for any $X,A$ in an $(\infty,1)$-topos $E$ we have $Hom_E(X,A) = Hom_{E/X}(1,X^*A) = \Gamma_{E/X}(X^*A)$. Recall that traditional “abelian” sheaf cohomology $H^n(X;A)$ is the case when $A$ is an $n$-fold delooping of a discrete abelian group object, and when $A$ is locally constant (whatever that means) it reduces to “cohomology with local coefficients” and further to the most traditional algebraic-topology sort of cohomology when $A$ is constant. Generalizing in a different direction, if $A$ is constant, not on a discrete abelian group, but on a spectrum, then we obtain classical “generalized cohomology”, and if we further generalize to spectrum objects in $E$ then we have “generalized sheaf cohomology” with coefficients in a sheaf of spectra. Note that $\Gamma$ preserves abelian group objects and spectrum objects, so that with these definitions, abelian cohomology theories always produce abelian objects.

In another thread I asked a question about *homology* from the nPOV (and David C kindly supplied some links to past discussions). A couple of answers were given, but I just thought of a slightly different way of stating it, which I like. Suppose our $(\infty,1)$-topos is locally ∞-connected, so that in addition to a right adjoint $\Gamma$, the constant stack functor $\Delta$ has a left adjoint $\Pi$. Now $\Pi$ won’t preserve abelian and spectrum objects, but by general “adjoint lifting theorems” I would nevertheless expect to be able to *construct* from it a functor $\Pi_{spectra}: Spectra(E) \to Spectra$ which is left adjoint to $\Delta_{spectra}: Spectra \to Spectra(E)$. It seems to me that it would make sense to regard $\Pi_{spectra}$ as a notion of “sheaf homology” with coefficients in a sheaf of spectra (perhaps a constant one).

It could be that this is way off-base, but I’m getting my intuition from the May-Sigurdsson theory of “parametrized” spectra, which should morally (I believe) be identifiable with “locally constant” sheaves of spectra over nice spaces. In their context, the pullback functor $r^* : Spectra \to Spectra/X$ always has both a left adjoint $r_!$ and a right adjoint $r_*$, and the left adjoint is homology while the right adjoint is cohomology. In particular, $r_! r^* M$, for a spectrum $M$, can be identified with the generalized homology theory $H_*(X;M)$.

Thoughts? Is this obviously true? Obviously false?

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