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Suppose that $X$ is a simplicial complex, and $f:X \rightarrow S^k$ a continuous map to a sphere. Is $f$ always homotopic to a simplicial map to the boundary of a $(n+1)$-simplex, $\partial \Delta^k$?

The simplicial approximation theorem implies that there is a barycentric subdivision of $X$, $X'$, and a simplicial map $g:X'\rightarrow \partial \Delta^k$ that is homotopic to $f$. The question is whether we can drop the barycentric subdivision in this particular case.

EDIT: this question originally asked "Does a nonzero cohomology imply existence of a simplicial map to a sphere?" Indeed, as Bertram Arnold pointed out, the complex projective plane has nonzero homology in degree 2, but has no non-null-homotopic maps to the 2-sphere.

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    $\begingroup$ Sorry, as you pointed out, $k$ was being used as an index and as a field. Now, k is supposed to be a field, and n is supposed to be the dimension. $\endgroup$
    – Quizzical
    Commented Jun 11, 2021 at 5:06
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    $\begingroup$ Not sure why you're saying that, if $H^n(X,k)$ is nontrivial, then there exists a non-null-homotopic map $X\rightarrow S^n$. Let $C_2$ be the group with 2 elements, and let $X$ be an Eilenberg-Mac Lane $K(C_2,1)$, for example, infinite-dim'l real projective space. Then $H^1(X,k)$ is certainly nonzero for many fields $k$ (after all, it is the group cohomology of the cyclic group $C_2$), but by the (proven) Sullivan conjecture, every pointed map from $X$ to $S^1$ is null-homotopic. Am I missing something? $\endgroup$
    – user164898
    Commented Jun 11, 2021 at 5:40
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    $\begingroup$ @A.S.: Maps from $X$ to $S^1$ is the first integral cohomology of the infinite dimensional real projective space. Calculating this is somewhat easier than proving the Sullivan conjecture. $\endgroup$
    – Oscar Randal-Williams
    Commented Jun 11, 2021 at 7:15
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    $\begingroup$ @OscarRandal-Williams: Yes, bringing up the Sullivan conjecture in that example is squashing a fly with a truck. I brought it up anyway, because it is an argument that generalizes to other finite groups and higher values of $n$, which I hope might help the original poster to identify what might be going on with the result that the question described. $\endgroup$
    – user164898
    Commented Jun 11, 2021 at 8:15
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    $\begingroup$ Another source of counterexamples to the claimed result is obtained by noting that the image of $f^*$ inside $H^n(X,k)$ must be annihilated by all cohomology operations (and the resulting secondary and higher cohomology operations). This boils down to lifting the map $X\to K(\mathbb Z,n)$ along $S^n\to K(\mathbb Z^n)$ using obstruction theory. The relevant obstruction groups involve the homotopy groups of $S^n$, so this will be essentially impossible to do in a systematic fashion. $\endgroup$
    – Bertram Arnold
    Commented Jun 11, 2021 at 8:44

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Take $X = \partial \Delta^3$. Then homotopy classes of maps from $|X|$ to $S^2$ correspond to elements of $\pi_2(S^2)\cong \mathbb{Z}$. But there are only finitely many maps of simplicial complexes $\partial \Delta^3 \to \partial \Delta^3$. So not all of them can be represented without further subdivision.

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