# Turning injection of homotopy groups to an isomorphism

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.

• 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? – Matt Feller Jun 16 '19 at 12:10
• Yes they induce injections on homotopies and homologies in MY situation. – user127776 Jun 16 '19 at 16:42
• A counterexample, which is even a map of infinite loop spaces and injective on homology, is $BSU\to BU$. – Charles Rezk Jun 16 '19 at 19:15

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.

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.

• The mapping cylinder. It is homotopy equivalent to the codomain. – Connor Malin Jun 16 '19 at 20:47
• 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 – Denis Nardin Jun 16 '19 at 20:51
• 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). – Mike Miller Jun 16 '19 at 21:12
• @DenisNardin I edited my answer to be more clear. – Connor Malin Jun 16 '19 at 21:20