An example of why this homomorphism of homotopy groups is not injective.

386 Views Asked by At

For a given path-connected space $X$, I recently learned that one could construct a CW complex $X_{1}$ by considering each generator $j: S^{q} \rightarrow X$ of $\pi_{q}(X)$ and setting $X_{1}=\bigvee_{(j,q)}S^{q}$, the wedge of spheres $S^{q}$, one for each generator $j$. The map $\gamma: X_{1} \rightarrow X$ defined by setting $\gamma$ to be $j$ on each $(j,g)th$ wedge component, induces a surjection of homotopy groups. This map however does not always induce an injection. I am having trouble understanding why this is so. Furthermore, I am unable to think of a specific example of this.

What are some examples of the spaces and the induced homomorphism above not being injective?

1

There are 1 best solutions below

0
On BEST ANSWER

The reason this shouldn't always induce an injection is there may be nontrivial relations between the elements of the homotopy groups that we're ignoring in $X_1$. For instance, carrying out this procedure for the torus, we get $X_1 = S^1 \vee S^1$, but the fundamental group of the torus is abelian (a fact ignored in $X_1$).

Another factor that leads to non-injectivity is that the homotopy groups of spheres can be (and often are) nontrivial in higher degrees, in ways that aren't reflected in $X$; the procedure for $\Bbb{CP}^\infty$ gives $X_1 = S^2$, but $\pi_n(\Bbb{CP}^\infty)=0$ for $n \neq 2$, and $\pi_3(S^2) = \Bbb Z$.

For, say, $\Bbb{RP}^2$ we get some of both worlds. Its fundamental group is $\Bbb Z/2\Bbb Z$, so twice the generator of $S^1$ is sent to zero (our $X_1$ can't notice torsion). But $X_1$ also has copies of $S^k$ for generators of $\pi_n(\Bbb{RP}^2) = \pi_n(S^2)$ for $n>1$, and these (many) $S^k$s have extra homotopy that's not reflected in $\Bbb{RP}^2$, etc.