Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Spanier's "Infinite Symmetric Products, Function Spaces and Duality," he makes the following claim:

Given some $X\hookrightarrow S^n$, and $X'$ which is an "$n$-dual" of $X$ (i.e. for some $k$, and all larger $k'$, $\Sigma^kX'\simeq\Sigma^k(S^n\setminus X)\simeq S^{n+k}\setminus X$, where the first equivalence is along a deformation retract), we wish to find a map $X\wedge X'\to S^{n-1}.$

To do so, we remove a point from $S^n\setminus(X\cup X')$ and so have $X,X'\hookrightarrow S^n\setminus\mathrm{pt.}\cong \mathbb{R}^{n}$. We define $$v:X\times X'\to S^{n-1},~(x,x')\mapsto\frac{x-x'}{\vert\vert x-x'\vert\vert}.$$

Spanier states that this map restricted to $X\vee X'$ is nullhomotopic under the condition that $X$ and $X'$ are connected (and the situation is in fact such that we only need to choose $X$ connected and we can get $X'$ to be connected by suspensions, by the first three part homotopy equivalence given). Spanier claims that this is related to the fact that $H^q(X\vee X')=0$ for every $q\geq n-1$.

The map on the smash product comes from shrinking the wedge to a point, obviously.

Thanks for any help on this, it's sort of a complicated, classical problem.

share|cite|improve this question
up vote 5 down vote accepted

We can answer this without anything fancier than cellular approximation. As user17786 points out, if the dimension of $X \vee X'$ is smaller than the dimension of the sphere $S^{n-1}$, then we win because we approximate the map $X \vee X' \rightarrow S^{n-1}$ by a cellular map, which lands in the $(n-2)$-skeleton of $S^{n-1}$, which is just the basepoint.

Moreover, we can suspend $X$ and $X'$ as many times as we like. If $X$ embeds into $S^n$ and its complement is homotopy equivalent to $X'$, then $\Sigma^l X$ embeds into $S^{n+k+l}$ and its complement is homotopy equivalent to $\Sigma^k X'$.

So, let $d$ be the dimension of the finite CW complex $X$ and let $d'$ be the dimension of $X'$. Then $\Sigma^k X \vee \Sigma^k X'$ has dimension max$(d,d')+k$, whereas $S^{n-1+k+k}$ is $(n+2k-2)$-connected. If $k$ is large then the second number is bigger than the first, and we win.

Now, can we answer the question without suspending $X$? Yes, if we allow ourselves to use Alexander duality, which tells us that $\tilde H^q(X) \cong \tilde H_{n-q-1}(X')$ and vice-versa. If $X$ is connected, then $\tilde H^0(X) = 0$ and $\tilde H_0(X) = 0$, so by Alexander duality, $H_{n-1}(X') = 0$ and $H^{n-1}(X') = 0$. (The higher homology and cohomology vanish as well.) Therefore $X'$ is homotopy equivalent to a complex of dimension at most $n-2$. (As the other user points out, this is proven by Hatcher in the section on minimal cell structures.) Anyway, if $X'$ is also connected, then the same goes for $X$, so $X \vee X'$ has dimension at most $n-2$ and we win again.

EDIT: This answer is a couple of years old but I need to correct an oversight made by past me. In this last paragraph I assumed $X$ and $X'$ are simply-connected, which is easy enough to achieve by suspending each one twice. If they are not simply-connected then there is an extra obstruction to being homotopy equivalent to a finite-dimensional complex: $H^n(X;G)$ must be zero for all twisted coefficient systems $G$. (See Wall's paper Finiteness conditions for CW-complexes.) I don't believe Alexander duality gives us anything with twisted coefficients - another good reason to just suspend and carry on with your life.

share|cite|improve this answer
Win win win! Win! – Aaron Mazel-Gee Nov 9 '11 at 18:46
I do believe you've got it! I'm nervous about changing the htpy type of X' by suspension since we want it to ultimately define a Spanier Whitehead dual to X, though this may be irrelevant. However your statement concerning duality I think is exactly the right one, and on fact the argument that Spanier seems to be using! Thanks! – Jon Beardsley Nov 10 '11 at 1:58
Spanier later generalizes his definition of duality to one that is stable, i.e. only needs to hold after X and X' have been suspended arbitrarily many times. When we pass from spaces to spectra, this becomes the notion of duality for spectra, totally analogous to duality of vector spaces. (Spheres are the "unit objects.") When two spectra are dual, the homology of one is the cohomology of the other (in any extraordinary homology or cohomology theory!) – Cary Nov 10 '11 at 6:40

I guess we can suppose that all the spaces are CW-complexes, hence finite, being subsets of a sphere.

If a CW-complex $X$ (your wedge) has dimension $\leq n-2$, then all maps to a $(n-2)$-connected $CW$-complex $Y$ (the sphere $S^{n-1}$) are homotopic to a constant (because $Y$ is homotopy equivalent to a complex $Z$ whose first nontrivial cell appears in dimension $n-1$, and then we can use the $CW$-approximation theorem for maps to deduce that all maps from $X$ to $Z$ map all cells to the trivial $0$-cell).

If $H^q(X)=0$ for all $q\geq n-1$, then by the universal coefficient theorem for cohomology (Theorem 3.2 in Hatcher's), $H_q(X)=0$ for all $q\geq n-1$ and $H_{n-2}(X)$ is a free abelian group. Now, by proposition 4C.1 in Hatcher's (on minimal cell structures), $X$ is homotopy equivalent to a $CW$-complex $Z$ of dimension at most $n-1$. Moreover, if $Z$ has a cell in dimension $n-1$ it is a 'relator', as explained there, and creates a torsion subgroup in $H_{n-2}(X)$, but this group was free, so $Z$ has dimension $\leq n-2$, and we can apply the first paragraph.

EDIT: (I summarize the discussion held in the comments) The above solution fails as $\pi_1(X)$ is required to be trivial for applying Proposition 4C.1. For assuring $X$ is simply connected, we have to suspend $X$ and change the dimension of the sphere where $X$ lives. Having done that, Cary's answer is simpler than this one. If we want to avoid suspending, i.e. we want to prove that ''for a space $X$ such that $H^q(X)=0$ for all $q\geq n$, the set of homotopy classes $[X,S^{n-1}]$ is trivial", I think it is better to work with the Postnikov tower of $S^{n-1}$. In this approach, it would be a direct consequence of Corollary 4.73 in Hatcher's: If $Y$ is a connected abelian $CW$-complex and $(W,X)$ is a $CW$-pair such that $H^{q+1}(W,X;\pi_q X)$ for all $q$, then every map $X\rightarrow Y$ can be extended to a map $W\rightarrow Y$. Take $W$ to be the cone of $X$ and $Y=S^{n-1}$, and note that $H^q(X)=0$ for all $q\geq n-1$ implies that $H^q(X;G)=0$ for all $q\geq n-1$ and for any (finitely generated) abelian group $G$.

share|cite|improve this answer
Sorry, yeah, everything is a finite CW-complex, or even some sub-polyhedron or something, if you like. – Jon Beardsley Nov 8 '11 at 23:12
One question, Hatcher requires $X$ to be simply-connected. Is this solved in this case by taking suspensions? – Jon Beardsley Nov 9 '11 at 0:41
I'm sorry, you are right: my answer requires that $\pi_1(X)$ has to be trivial. A better approach is to use the Postnikov tower of $Y$ instead of the skeleton of $X$, since cohomology groups describe maps to Eilenberg-Mac Lane spaces and the Postnikov tower of $S^{n-1}$ will be easier to handle than the skeleton of an arbitrary subcoMplex of $S^{n-1}$. – user17786 Nov 9 '11 at 9:46
In this case the answer is given by Corollary 4.73 in Hatcher's: If $Y$ is a connected abelian $CW$-complex and $(W,X)$ is a $CW$-pair such that $H^{q+1}(W,X;\pi_q X)$ for all $q$, then every map $X\rightarrow Y$ can be extended to a map $W\rightarrow Y$. Take $W$ to be the cone of $X$, and note that $H^q(X)=0$ for all $q\geq n-1$ implies that $H^q(X;G)=0$ for all $q\geq n-1$ and for any finitely generated abelian group $G$. – user17786 Nov 9 '11 at 9:46
"In the meantime, we have the complement getting embedded in higher and higher dimensional spheres, so that we have to assume that at some point it becomes contractible?". I don`t see how. The complement is always the same (embedded in different spheres), so its homotopy type doesn't change (as you say first). – user17786 Nov 9 '11 at 16:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.