# Functoriality of the Fundamental group

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.

• 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. – Chris Eagle May 15 '13 at 16:43
• 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. – bbnkttp May 15 '13 at 16:52
• 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. – Zhen Lin May 15 '13 at 17:02
• @MohamedHashi: How can you be satisfied with the construction of the objects when you haven't given a construction of objects? – Chris Eagle May 15 '13 at 17:04
• 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. – bbnkttp May 15 '13 at 17:11

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.
• 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. – Martin Brandenburg May 15 '13 at 20:07
• What if we take, for example, $G=S^1$?. Then $BG = K(\mathbb{Z},2)$. $BG$ is not always an E-M space – Drew May 16 '13 at 0:30
• 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. – Dan Ramras May 16 '13 at 0:52
• @Drew As is clear from the construction of $B G$ I give, I am only discussing discrete groups. – Zhen Lin May 16 '13 at 7:09