# Constant maps from simplicial complexes to the n-sphere $S^n$

I've come unstuck and the following and was hoping someone could provide advice:

X a space, triangulable as a simplicial complex with no n-simplices. Prove any cts map: $X \to S^n$ is homotopic to a constant map.

Now the only theorem I know (at undergrad level!) which would be any use here is the simplicial approximation theorem, which I suspect is what's wanted of me. However, I haven't had much cause to use it before and so I don't have the best understanding of it; I'm not fond of working with simplices. It certainly isn't at all clear to me how we would make use of the fact that we can triangulate as a simplicial complex without any n-simplices; it makes me think that X is some how 'less than n-dimensional' in some sense, and then perhaps somehow using the fact that $S^n$ is simply connected we might be able to obtain a homotopy to the constant map via some triangulation of $S^n$ and then from the triangulation to $S^n$ itself, which must then be trivial (similar to the 'maps between $\mathbb{R}^n$ and $\mathbb{R}^m$' result). However, I'm just mentioning whatever comes into my head here, it's not really any sort of formalised idea, and most of it is likely to be wrong. Could anyone help?

Thanks very much - M

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Two hints: 1) Can you see how to prove this if you can assume that $f : X \to S^n$ is homotopic to a map which avoids at least one point? 2) Can you see how to show that $f$ is homotopic to such a map? (You are correct that you should use simplicial approximation here.)
You don't need to use the fact that $S^n$ is simply connected (for $n > 1$); this is in fact one way to prove this.
@El Moro: this is impossible since any map $f : X \to S^n$ as in the question induces the zero map on $H_n$. – Qiaochu Yuan Apr 30 '11 at 18:19
Ah, of course not, since their boundaries would eventually have $n$-simplices – Dan Ramras May 1 '11 at 5:11