# Is every map from a simplicial complex to a sphere homotopic to a simplicial map to the boundary of a k-simplex?

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.

• 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. Jun 11 at 5:06
• 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?
– A.S.
Jun 11 at 5:40
• @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. Jun 11 at 7:15
• @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.
– A.S.
Jun 11 at 8:15
• 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. Jun 11 at 8:44

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.