I originally posted this on MSE but didn't get much of a response, so I'll attempt to post it here. Let me know if this is not appropriate.


Let $M$ be a smooth manifold of dimension $n$. Let $\alpha \in \pi\_i (M)$, for some $i \leq n$, and $f:S^i \to M$ be a map of the sphere representing the homotopy class $\alpha$. Can $f$ always be chosen to be an immersion? I.e. does every homotopy class of the maps from the sphere $S^i$ to $M$ contain an immersion?


In the comments of my [original post](https://math.stackexchange.com/questions/3895872/can-every-element-of-a-homotopy-group-of-a-smooth-manifold-be-represented-by-an), it is claimed that it doesn't always work when $i=n$, and that it should work for small enough $i$. What about for other values in between?



Can every element of a homotopy group of a smooth manifold be represented by an immersion?
There exists a simply-connected closed $6$-manifold $M$ with a homotopy class $\alpha\in \pi\_4(M)$ which does not contain an immersion. The following argument is due to Diarmuid Crowley, after we realized that my argument with Stiefel-Whitney classes could not produce examples.


According to Wall,


*Wall, C. T. C.*, [**Classification problems in differential topology. V: On certain 6- manifolds**](http://dx.doi.org/10.1007/BF01389738), Invent. Math. 1, 355-374 (1966); Corrigendum. Ibid. 2, 306 (1967). [ZBL0149.20601](https://zbmath.org/?q=an:0149.20601)


there exists a $1$-connected closed $6$-manifold $M$ such that $H^\*(M)\cong H^\*(\mathbb{C}P^3)$, the cup product $H^2(M)\times H^2(M)\to H^4(M)$ is trivial, and the first Pontryagin class $p\_1(M)\in H^4(M)$ is non-zero. The relevant classification result is Theorem 5 in the linked paper.


**Claim:** The Hurewicz map $\pi\_4(M)\to H\_4(M)$ is onto.


**Proof:** $M$ has the homotopy type of a CW-complex with one cell in each even dimension betoween $0$ and $6$. Since the cup square of the generator in $H^2(M)$ is trivial, it follows that the attaching map $S^3\to S^2$ of the $4$-cell has trivial Hopf invariant, therefore is trivial. So $M^{(4)}\simeq S^2\vee S^4$. The claim follows.


Now let $\alpha\in \pi\_4(M)$ be a class whose Hurewicz image is a generator $x\in H\_4(M)$. Suppose $\alpha$ is represented by an immersion $f:S^4\looparrowright M$. The normal bundle of $f$ is a rank $2$ bundle $\nu(f)$ over $S^4$, hence is trivial. Also, $TS^4$ is stably trivial. The isomrphism of bundles $\nu(f)\oplus TS^4 \cong f^\*(TM)$ and naturality of Potryagin classes now implies that $f^\*(p\_1(M))=0\in H^4(S^4)$. Thus
$$
0=\langle f^\*(p\_1(M)),[S^4]\rangle = \langle p\_1(M),f\_\*[S^4]\rangle = \langle p\_1(M),x\rangle
$$ which implies $p\_1(M)=0$, a contradiction.


