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My definitions of Bordism are from Tom Dieck's Algebraic topology book. Briefly, a singular manifold, $M \xrightarrow{f} pt $, for a closed smooth oriented manifold $(M, \omega)$ without boundary is nullbordant, inside a point, if it is the oriented restriction of a map $B \xrightarrow{F} pt$ to its boundary: Here $B$ is a closed oriented smooth manifold( with boundary) an orientation preserving diffeomorphism $\phi: M \to \partial B$ and the condition is that $F\circ \phi=f$.

One defines two oriented singular manifolds $M_1 \xrightarrow{f_1} pt$, $M_2 \xrightarrow{f_2} pt$ to be cobordant, inside a point if the map defined on $M_1 \sqcup -M_2 \to pt$ defined on $M_1$ to be $f_1$ and on $-M_2$ to be $f \circ (-M \to M)$, is nullbordant.

I don't understand why $\Omega^*_n(pt)$ the equivalence classes of singular n-dimensional oriented manifolds(i.e. maps from smooth n-dimensional oriented manifolds) inside a point, are not nonzero.

A well known consequence of these definitions, which I formally understand, by using the pontryagin thom map, is that these classes are in correspondence with $MSO(pt^+):=[MSO(n), S^0]_+$,( where $MSO(n)$ is the thom space of the tautological bundle over $BSO(n)$). $MSO(n)$ is a connected(even (n-1)-connected )space. It seems to me that there is only one basepoint preserving map to $S^0$, and thus only one homotopy class of basepoint preserving maps to $S^0$.

Why then are the groups $\Omega^*_n(pt)$ nonzero?

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The oriented cobordism group are the homotopy groups of $MSO$, not the cohomotopy groups as you've written. And the indexing isn't right-you want the stable homotopy of the spectrum $MSO$, so $\varinjlim \pi_{n+k} MSO(n)=\Omega_k$.

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  • $\begingroup$ I understand now. The oriented cobordism groups, just like any cobordism groups with structure have the following correspondence: $\Omega_n(X)=colim_k \pi_{n+k}(M(id_x X \times \xi(BSO(k))))$. When $X$ is a point then this smash product becomes $colim_k \pi_{n+k}(MSO(k))$. $\endgroup$ – user062295 Apr 13 '16 at 1:31
  • $\begingroup$ In particular, using the fact that for $n=0$ this is $\pi_2(BSO(2))$ by the hurewicz theorem and thom isomorphism theorem, and that $\pi_2(BSO(2))=\pi_2(CP^\infty)=\Z$ we see that $\Omega_0(pt)=\mathbb{Z}$ $\endgroup$ – user062295 Apr 13 '16 at 1:35
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Let $X$ be a topological space. Two oriented singular manifolds $f_1 : M_1 \to X$, $f_2 : M_2 \to X$ are orientedly cobordant inside $X$ if:

  • $M_1$ and $M_2$ are orientedly cobordant (there is $B$ with $\partial B = M_1\sqcup -M_2$), and
  • there is a map $f : B \to X$ such that $f|_{M_1} = f_1$ and $f|_{M_2} = f_2$.

In the case that $X$ is a point, $f_1 : M_1 \to X$ and $f_2 : M_2 \to X$ are orientedly cobordant if and only if $M_1$ and $M_2$ are orientedly cobordant; that is, $\Omega_n^*(pt) \cong \Omega^{SO}_n$. This is because the second condition always holds in this case if the first one does. More precisely, if $B$ is such that $\partial B = M_1\sqcup -M_2$, then the unique map $f : B \to X$ necessarily restricts to the correct maps on the boundary components.

Given the above, to show that $\Omega^*(pt)$ is non-zero, it is enough to give an example of a closed oriented manifold which is not null-cobordant. The simplest such example occurs in dimension zero: a point.

You might find this useful reading when it comes to oriented (co)bordism.

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