12
$\begingroup$

The fundamental group is a functor from the category of pointed topological spaces to the category of groups.

Therefore every base-point preserving continuous function $f$ between pointed topological spaces induces a homomorphism $f_*$ between the fundamental groups. This is done by composing the loops with $f$, which is well-defined, because homotopy is also preserved under $f$.

Can we switch this around?

Every group is the fundamental group of a CW-complex, which can be constructed according to how many generators and relations the group has.

Can a continuous function be constructed for every homomorphism such that the continuous function induces the homomorphism? If the fundamental group functor is 'surjective', one has a pre-image at least.

How do you go from the algebraic to the topological with the morphisms? I have no idea.

$\endgroup$
  • 3
    $\begingroup$ Before you can even ask your question, you need to pick a canonical CW complex for each group. The only way I can see to do this canonically is using everything in the group as a generator and every possible relation. $\endgroup$ – Chris Eagle May 15 '13 at 16:43
  • 1
    $\begingroup$ I understand that this gives problems, just as the fact that CW complexen don't have to be homeomorphic to each other to have the same fundamental group(add a n-cell, with $n\geq 3$). But I was already satisfied with the reversed construction of the objects(from group to topo. space). Doing this for mappings is the problem. $\endgroup$ – bbnkttp May 15 '13 at 16:52
  • 3
    $\begingroup$ Aren't you basically asking whether there is a choice of Eilenberg–MacLane spaces $K(G, 1)$ that is functorial in $G$? The answer is yes. $\endgroup$ – Zhen Lin May 15 '13 at 17:02
  • $\begingroup$ @MohamedHashi: How can you be satisfied with the construction of the objects when you haven't given a construction of objects? $\endgroup$ – Chris Eagle May 15 '13 at 17:04
  • $\begingroup$ I am satisfied with, given a group G, being able to construct a topological space with a fundamental group isomorphic to G. I realise that this is not extendible to a well defined mapping from $Grp$ to $Top$. But having realised this, I would like to know what can be done on the morphisms side of the story, which is not mentioned in Hatcher's for example. $\endgroup$ – bbnkttp May 15 '13 at 17:11
6
$\begingroup$

Let $B : \mathbf{Grp} \to \mathbf{Top}_*$ be the functor obtained by defining $B G$ to be the geometric realisation of the nerve of $G$ (considered as a 1-object category), i.e. the simplicial set $$\cdots \mathrel{\lower{0.5ex}{\begin{array}{c} \smash{\to} \\ \smash{\to} \\ \smash{\to} \\ \smash{\to} \end{array}}} G \times G \mathrel{\lower{0.5ex}{\begin{array}{c} \smash{\to} \\ \smash{\to} \\ \smash{\to} \end{array}}} G \rightrightarrows 1$$ where the degeneracies maps insert the unit element at the appropriate location and the face maps compose adjacent pairs of elements.

It is well-known that $B G$ is a $K (G, 1)$ Eilenberg–MacLane space, i.e. $B G$ is a path-connected topological space such that $\pi_1 (B G, *) \cong G$ and $\pi_n (B G, *) = 1$ for all $n > 1$. Moreover, the isomorphism $\pi_1 (B G, *) \cong G$ is induced by the obvious correspondence: send each element of $G$ to the loop in $B G$ that realises the corresponding 1-simplex in the nerve. It follows that $\pi_1 \circ B$ is naturally isomorphic to $\mathrm{id}_{\mathbf{Grp}}$ as a functor.

$\endgroup$
  • 1
    $\begingroup$ Does this really answer the question? If $F \circ G \cong \mathrm{id}$, then $\hom(x,y) \to \hom(G(x),G(y)) \to \hom(F(G(x)),F(G(y))$ is bijective, hence $\hom(a,b) \to \hom(F(a),F(b))$ is surjective provided that $a,b$ are in the image of $G$. Not every space is an Eilenberg-MacLane space. $\endgroup$ – Martin Brandenburg May 15 '13 at 20:07
  • $\begingroup$ What if we take, for example, $G=S^1$?. Then $BG = K(\mathbb{Z},2)$. $BG$ is not always an E-M space $\endgroup$ – Drew May 16 '13 at 0:30
  • 1
    $\begingroup$ Drew, I think the question was about discrete groups, not topological ones. Also, usually E-M space just means a space with exactly at most non-zero homotopy group, so K(Z,2) is an E-M space. $\endgroup$ – Dan Ramras May 16 '13 at 0:50
  • 3
    $\begingroup$ To put what Zhen Lin wrote in more explicit form, given a group homomorphism $f: G\to H$, there is an induced map $Bf: BG\to BH$. It is elementary to check that $(Bf)_*: \pi_1 BG\to \pi_1 BH$ agrees with $f$ under the canonical isomorphisms $G\simeq \pi_1 BG$ and $H\simeq \pi_1 BH$. It seems to me that this is all the question asked for. $\endgroup$ – Dan Ramras May 16 '13 at 0:52
  • $\begingroup$ @Drew As is clear from the construction of $B G$ I give, I am only discussing discrete groups. $\endgroup$ – Zhen Lin May 16 '13 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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