1:09 AM
@ACuriousMind Does a homotopy equivalence induce isomorphisms on the direct limit of homotopy groups in all dimensions?
@ACuriousMind So the directed set here is the sequence $a_1<a_2<\cdots$
And the groups are $\pi_n(K_i)$?
So what are the homomorphisms $f_{ji}:\pi_n(K_i)\to\pi_n(K_j)$?
I don't know how this is a direct system of groups
@ACuriousMind First, let's go back to the general case
$D$ is a directed set and $G_\alpha$ are abelian groups for $\alpha\in D$. I think the homomorphisms $f_{\beta\alpha}:G_\alpha\to G_\beta$ tell us how they "relate" to each other
the condition $f_{\gamma\beta}\circ f_{\beta\alpha}=f_{\gamma\alpha}$ basically tells us that if we know how $G_\alpha$ relates to $G_\beta$ and how $G_\beta$ relates to $G_\gamma$, then we know how $G_\alpha$ relates to $G_\gamma$
Now, we have the equivalence relation $f_{\beta\alpha}(g)\sim g$ $\forall g\in G_\alpha$ $\forall \beta>a$.
right, so what this does is it tells us that the $g'(:= f_{\beta\alpha}(g))$ in $G_\beta$ related to $g\in G_\alpha$ is basically "the same"
and that all $g'$s in all the "higher" groups are basically "the same"
So, this fixes a relationship between all the groups and between all the elements in all the groups
the maps need not be surjective
ah, but they tell you how the elements in each group are related to the one right above, and so on
Now, we define $\varinjlim G_\alpha=(\bigoplus G_\alpha)/\sim$.
So, we put together all the groups and identify the elements which we said are "the same"
@ACuriousMind Ok. So are the homomorphisms in the Milnor case just maps induced on the homotopy groups by the inclusion maps?
Do I sound insane right now
@ACuriousMind So, we have these nested sets $K_1\subset K_2\subset\cdots$
Now, we have to take the union $\bigcup K_i$, presumably
Now, what is $\pi_n\left(\bigcup K_i\right)$