Hurewicz map factors through bordism homology I've read in multiple sources that the hurewicz map $h \colon \pi_n(X) \to H_n(X)$ factors through oriented bordism homology.
I'm particularly interested in the injectivity of the map $h \colon \pi_n(X) \to MO_*(X)$. Is this only given if X is $(n-1)$-connected? Can you give a source where this is proven?
Best regards
Thorben 
 A: I have proven the injectivity of the Hurewicz map $h \colon \pi_n(x) \to MO_*(X)$ where $X$ is (n-1)-connected for $n \geq 2$.
Let $f$ be in the kernel of the Hurewicz map i.e there exists a manifold $W$ together with a map from $W$ to $X$ such that $\partial W=S^n$ and such that $g|_{\partial W}=f$.
A simple argument about homotopy theory shows us that we can homotop $g$ to be constant on every handle of degree $k$ if $k \leq n-1$. So what about the $(n-1)$-handles. Let $W'$ denote $W$ without $n$-handles and $(n+1)$-handles and $\Phi \colon S^{n-1}\times D^1 \to \partial W'$ denote an embedding, which characterizes the first $n$-handle (If there is no $n$-handle we can skip this part). Fix a point $p$ in $S^{n-1}$. Consider first the case, where a path $\gamma \colon [0,1] \to \partial W' - S^{n-1} \times (0,1)$ exists such that $\gamma (0) = (p,0)$ and $\gamma (1) = (p,1)$ and such that $\gamma$ together with the embedded arc $\{p\} \times [0,1]$ forms an embedded $S^1$. Note that only the first part is to check and the second part can be arranged for every such path. 
            Since $W$ is orientable this embedded $S^1$ has an embedding of its trivial tubular neighbourhood $\Psi \colon S^1 \times D^{n-1} \to \partial W'$. Note that $\partial (W' +(\Phi^n))$ is isomorphic to the boundary of $\partial (W + (\gamma))$. This can be seen since from the view of the boundary $S^{n-1} \times D^1$ is just a $1$-handle and what happens if we glue in $\Phi$? We delete it and glue in two discs to fill the holes. But now our $\Psi$ fullfils the conditions from the cancelation lemma to erase this $1$-handle and therefore we conclude that $\partial(W+(\Phi))$ and $\partial(W'+(\Psi)$ are actually isomorphic. Note here that $W'+ (\Phi^n)$ and $W' + (\Psi^2)$are not isomorphic! Since $X$ is at least simply connected we can make $g$ constant on $\Psi$ and we conclude that we can actually extend $g$ onto $W' + (\Psi^2)$. Therefore we conclude that we can choose $W$ without loss of generality to not have such handles. 
Let now $\tilde{W}$ denote $W$ without $(n+1)$-handles and the last $n$-handle $(\Phi^n)$.
            Since $\partial \tilde{W} - S^{n-1} \times (0,1)$ is non-path-connected we get that $\partial \tilde{W} + (\Phi^{n+1})$ is again not connected.
            But now we get that since $\partial W$ is connected, that in at least one of these two components there needs to be glued in a disk $D^{n+1}$.
            Let $\Psi$ denote the first embedding such that $\partial (\Phi^n) \cap \mathrm{Im}(\Psi^{n+1}) \neq \emptyset$. 
            But now since $S^n$ is compact and of dimension $n$ we get that the embedding in $\partial \tilde{W} + (\Phi^{n+1})$ is a diffeomorphism onto a connected component of $\partial \tilde{W} +(\Phi^{n+1})$. The cancellation lemma tells us now that $\Phi^n$ and $\Psi^{n+1}$ cancel each other out.
Therefore we conclude that we can homotop $g$ such that it is constant on $W$ without the $(n+1)$-handles but since the $(n+1)$-handles just fill parts of the boundary we actually conclude that our map can be homotoped to be constant on the boundary of $W$. Therefore $h$ is injective.
