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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.

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    @Yuan by the way when we have a surjective function what are the conditions to fulfill in order to have a continuous section?2011-04-30
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    @El Moro: this doesn't seem particularly relevant to the current question; why don't you ask it as a separate question?2011-04-30
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    @Yuan I was thinking about what it would mean to have a continuous function from the n-sphere to X with no n-simplices2011-04-30
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    @El Moro: this is impossible since any map $f : X \to S^n$ as in the question induces the zero map on $H_n$.2011-04-30
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    Yeah that is the conclusion. So based on your 1st hint and if we have the conditions to have a section from the n sphere to X that has no n simplex. could we show that something is going wrong?2011-04-30
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    @El Moro: I'm not sure I follow. What does the existence of a section have to do with the question the OP is asking?2011-04-30
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    The existence of a continuous section would lead to a subset of X being homeomorphic to the n-sphere. and I think that might lead to a contradiction.. so we end up saying that the function f is not surjective and here we'll use your first hint.2011-04-30
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    @El Moro: I'm still not sure I follow. It _is_ in fact possible for $f$ to be surjective; one can construct an appropriate generalization of a space-filling curve. The point is that nevertheless $f$ is homotopic to a map which is not surjective.2011-04-30
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    Aha fine now :)2011-04-30
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    @Qiaochu - 1) I suppose we could use the fact that an n-sphere minus a point is homeomorphic to $\mathbb{R}^n$: so if we use that it reduces to showing that any such map to $\mathbb{R}^n$ must be constant: but then any map to to $\mathbb{R}^n$ will be nullhomotopic, right? So then if that's correct, it just remains to show 2) - but I haven't had much of an opportunity to make use of the SAT before so I'm afraid I can't see how to use it to show that f must be homotopic to such a map. We know that between simplicial complexes we have a simplicial approximation to any such f, but not sure here.2011-04-30
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    Also, I haven't used the fact it has no n-simplices.2011-04-30
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    @Matt: step 2 is where you use the fact that there are no $n$-simplices. What can you say about a simplicial approximation to $f$, keeping this in mind?2011-04-30
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    @Qiaochu: Well if g is any simplicial approximation to f (so |g| is homotopic to f), then g is defined on some sufficiently large barycentric subdivision of K (where |K|=X) and then extended linearly - since K has no n-simplices, we certainly know that K$^{(r)}$ has no n-simplices (where K$^{(r)}$ denotes the r-th subdivision); but can we say anything more than this? Since the number of vertices in each barycentric division will obviously increase at each stage I'm not sure that's any use, so I'm afraid I don't see what you're driving at, sorry.2011-05-01
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    @Matt: yes, you know that no barycentric subdivision of $X$ has $n$-simplices. Together with the fact that $g$ is simplicial, what does that tell you about the image of $g$ in $S^n$?2011-05-01
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    Are we allowing simplices of dimension greater than $n$ here?2011-05-01
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    Ah, of course not, since their boundaries would eventually have $n$-simplices2011-05-01