Does $\pi_1$ have a right adjoint? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:30:41Z http://mathoverflow.net/feeds/question/45351 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45351/does-pi-1-have-a-right-adjoint Does $\pi_1$ have a right adjoint? tetrapharmakon 2010-11-08T20:13:18Z 2010-11-08T22:30:38Z <p>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 <em>counity</em>.</p> <p>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 <em>unity</em> 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.</p> <p>The key point, quite trivial, to answer is: it is well known that an adjunction is uniquely determined by one among <em>unity</em> and <em>counity</em>, provided the one is universal. But $\boldsymbol\varepsilon\colon \pi_1\circ\mathcal K\to \mathbf{1}_{\mathbf{Grp}}$ is an <em>equivalence</em>: can I conclude that it is universal?</p> http://mathoverflow.net/questions/45351/does-pi-1-have-a-right-adjoint/45361#45361 Answer by Neil Strickland for Does $\pi_1$ have a right adjoint? Neil Strickland 2010-11-08T20:41:30Z 2010-11-08T20:41:30Z <p>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. </p> http://mathoverflow.net/questions/45351/does-pi-1-have-a-right-adjoint/45375#45375 Answer by Jeremy Brazas for Does $\pi_1$ have a right adjoint? Jeremy Brazas 2010-11-08T22:30:38Z 2010-11-08T22:30:38Z <p>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.</p>