MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Eilenberg and Mac Lane showed that given a group $G$ there exists a pointed topological space $X_G$ such that $\pi(X_G,\bullet)\cong G$. It is obviously a way to "invert direction" to the functor $\pi_1\colon \mathbf{Top}^\bullet\to \mathbf{Grp}$ to a functor $\mathcal K\colon \mathbf{Grp}\to \mathbf{Top}^\bullet$ such that $\pi_1(\mathcal K(G),\bullet)\cong G$ (almost by definition). This is equivalent to say that there exists a natural transformation (equivalence, in this case) between $\pi_1\circ\mathcal K$ and $\mathbf{1}_{\mathbf{Grp}}$, which turns out to resemble some sort of counity.

It would be wonderful if I could define an adjunction between the two categories in exam, given by the two functors. Everytime I try to think about some sort of unity to this hypotetical adjunction I poorly fail: considering the vast literature in the field of algebraic topology, I believe in only two possible cases. The first, nothing interesting arises from this adjunction. The second, there is no sort of adjunction.

The key point, quite trivial, to answer is: it is well known that an adjunction is uniquely determined by one among unity and counity, provided the one is universal. But $\boldsymbol\varepsilon\colon \pi_1\circ\mathcal K\to \mathbf{1}_{\mathbf{Grp}}$ is an equivalence: can I conclude that it is universal?

share|cite|improve this question
up vote 11 down vote accepted

In the more general setting the answer is no. Left adjoints preserve colimits and it is not true that $\pi_{1}(X\vee Y)\cong \pi_{1}(X)\ast\pi_{1}(Y)$ for all spaces $X,Y$ (even compact metric spaces). For instance, if $(\mathbb{HE},x)$ is the usual Hawaiian earring, let $X=Y=C\mathbb{HE}=\mathbb{HE}\times I/\mathbb{HE}\times\{1\}$ be the cone on the Hawaiian earring with basepoint the image of $(x,0)$ in the quotient. It is a theorem of Griffiths that $\pi_{1}(C\mathbb{HE}\vee C\mathbb{HE})$ is uncountable and not the free product of trivial groups.

share|cite|improve this answer
Clear, clear. Thanks. :) – Fosco Loregian Nov 9 '10 at 0:05

Let $\mathcal{C}$ be the homotopy category of connected pointed CW complexes. Then the functor $\pi_1:\mathcal{C}\to\mathbf{Grp}$ does indeed have a right adjoint. The cleanest way to define it is to use the simplicial classifying space functor $B:\mathbf{Grp}\to\mathcal{C}$. For the unit of the adjunction you need a natural homotopy class of maps $X\to B(\pi_1(X))$. This can be done by obstruction theory, by induction up the skeleta of $X$. Alternatively, one can define a more canonical map from $|SX|$ (the geometric realisation of the singular complex) to $B(\pi_1(X))$ by simplicial methods, and then use the fact that there is a natural map $|SX|\to X$ which is a homotopy equivalence when $X\in\mathcal{C}$. The first place I would look for more details would be Peter May's old book on simplicial objects in algebraic topology.

share|cite|improve this answer
Goerss-Jardine is probably going to be more useful than Peter May's book, which is really outdated. – Harry Gindi Nov 8 '10 at 21:21
That copy of the book is copyright-infringing (not that I care), and it's probably better if you can avoid posting it on MO. – Harry Gindi Nov 9 '10 at 0:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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