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Assume we have a connected CW-complex $Y$ and $X\hookrightarrow Y$ a connected sub-complex. We know that the inclusion induces injection on all homotopy groups. Is it true (or under what conditions it can be true) that we can attach cells to $Y$ so that the inclusion induces isomorphism on homotopy groups? (This will subsequently imply that $X$ is a deformation retract of the enlarged $Y$.)

You can further more assume that the inclusion of $X$ is a map of infinite loop spaces. If that helps. Edit: The inclusion is also injective on the homologies.

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  • $\begingroup$ It is not true that an inclusion of CW complexes induces an injection of fundamental groups. For example, $S^1\hookrightarrow D^1$. So when you say ‘we know’ are you adding it as another assumption? $\endgroup$ – Matt Feller Jun 16 at 12:10
  • $\begingroup$ Yes they induce injections on homotopies and homologies in MY situation. $\endgroup$ – user127776 Jun 16 at 16:42
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    $\begingroup$ A counterexample, which is even a map of infinite loop spaces and injective on homology, is $BSU\to BU$. $\endgroup$ – Charles Rezk Jun 16 at 19:15
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Your question is equivalent to the following:

Given a cellular inclusion $i : X\to Y$, when is there a retraction $r:Y \to X$?

(Being a retraction means that $r\circ i: X\to X$ is the identity.)

The answer is usually phrased in terms of obstruction theory.

For simplicity, let's assume that $Y$ is a finite complex obtained from $X$ by attaching a single $j$-cell, i.e., $Y = X \cup_f D^j$, where $f: S^{j-1} \to X$ is the attaching map. Assume also that $X$ is a based space an $f$ is a based map.

In this case, it is easy to check that the desired retraction $r: Y \to X$ exists if and only if the homotopy class $[f] \in \pi_{j-1}(X)$ vanishes. We can think of this class as an obstruction lying in $$ \theta \in H^j(Y,X;\pi_{j-1}(X)) $$ (the $j$-cohomology group of the pair $(Y,X)$ with coefficients in $\pi_{j-1}(X)$).

Now, in the general case, we inductively assume that a retraction $$r_{j-1}: X_j \cup_{X_{j-1}} Y_{j-1}\to X$$ has already been specified where $Y_{j-1}$ is $(j-1)$-skeleton of $Y$. We wish to extend the retraction to $Y_j$. For every cell of $Y_j$ that is not lying in $X$, we have an obstruction in $\pi_{j-1}(X)$ defined as above. If we vary the cells, we obtain an element of $$ H^j(Y_j,Y_{j-1} \cup X_j ;\pi_{j-1}(X)) $$ whose vanishing is both necessary and sufficient to finding an extension $r_j: X \cup Y_{j} \to X$. Notice that the displayed cohomology group is the cellular $j$-cochains of the pair $(Y,X)$ with coefficients in $\pi_{j-1}(X)$. It turns out that the element in question is a cocycle in this cochain complex.

However, notice we made a choice: suppose we had used a different $r_{j-1}$?

Then the obstruction can change. With a little effort one can eventually see that the obstruction changes by a coboundary. So if we take into account all the choices, the cocycle is defined only up to a coboundary.

The upshot: there is a sequence of obstructions $$ \theta_j \in H^j(Y,X;\pi_{j-1}(X)) $$ such that $\theta_j$ is defined when $\theta_{j-1}$ vanishes. Furthermore all the obstructions vanish iff a retraction $Y\to X$ exists.

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Consider $i: S^1 \hookrightarrow M_f$ where $f:S^1 \rightarrow S^1$ is squaring and $M_f$ denotes the mapping cylinder. Then the inclusion induces a map $\pi_1(S^1) \rightarrow \pi_1(M_f)$ which has image $2 \mathbb{Z}$. Adding additional segments and attaching disks is the same as adding generators and relations to the presentation $\langle x,y | y=2x \rangle$ and the goal is to end up with $y$ generating the entire group with $|y|=\infty$. This means that we must have $x=ny=2nx$ which implies $x$ has finite order which in turn implies $y$ has finite order. This means that we cannot attach segments and disks to make the inclusion induce an isomorphism.

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  • $\begingroup$ The mapping cylinder. It is homotopy equivalent to the codomain. $\endgroup$ – Connor Malin Jun 16 at 20:47
  • $\begingroup$ Sorry for the comments, now I got what you mean (you are basically taking a cofibrant replacement of the squaring map). You might want to explain a bit the notation though, because it gave me pause for one moment $\endgroup$ – Denis Nardin Jun 16 at 20:51
  • $\begingroup$ you can forget the notation completely by remembering that $M_f$ is the Mobius band and the map from the circle is the inclusion of the boundary (but of course, the notion of mapping cylinder is independently useful). $\endgroup$ – Mike Miller Jun 16 at 21:12
  • $\begingroup$ @DenisNardin I edited my answer to be more clear. $\endgroup$ – Connor Malin Jun 16 at 21:20

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