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So taking the fundamental group of a pointed topological space can be thought of as applying a functor, $\pi_1$, that gives you something in the category of groups. Does $\pi_1$ have an adjoint?

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  • $\begingroup$ First of all, to make $\pi_1$ a functor you probably need to have it operate on the category of topological spaces with a base point. $\endgroup$ – Daniel Schepler Jun 16 '17 at 19:23
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    $\begingroup$ Hint for left adjoint: $(S^1, (1, 0))$ is an equalizer of two functions $(\mathbb{R}^2, (1, 0)) \to (\mathbb{R}, 1)$. Does the $\pi_1$ functor preserve this equalizer? $\endgroup$ – Daniel Schepler Jun 16 '17 at 19:24
  • $\begingroup$ Right, otherwise we would land in the category of groupoids. I'll edit to make that clear. $\endgroup$ – Bob Jun 16 '17 at 19:24
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    $\begingroup$ And for the right adjoint: I think $S^1$ is also a pointed coequalizer of two maps $(\{ 0, 1 \}, 0) \to ([-1, 1], 0)$. $\endgroup$ – Daniel Schepler Jun 16 '17 at 19:43
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If $\pi_1: Top_* \to Groups$ is a left adjoint of something, then it would preserve all colimits, which it does not, since we need some openness conditions for example. If it is a right adjoint of something then it preserves all limits, but it does not preserve, for example, pullbacks.

This is what makes the Seifert-Van Kampen Theorem somewhat magical, calculating in some circumstances, a non abelian invariant.

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If you see $\pi_1$ as a functor from the homotopy category of pointed connected CW-complexes to the category of groups, it is a left adjoint (that does not contradict Ronnie Brown's answer). See https://mathoverflow.net/q/109779/24563 or https://mathoverflow.net/a/45361/24563. The right adjoint is the classifying space functor.

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  • $\begingroup$ Thanks, you linked interesting reads. I'm also interested in the following iso from the first article you linked $\text{Hom}(\pi_1(X), G) \cong [X,K(G,1)]$. I couldn't find the reference in Hatcher's text. 1B.9 is a different theorem. Is that a typo or did Hatcher take that out in a later version? $\endgroup$ – Bob Jun 27 '17 at 5:31
  • $\begingroup$ For me, it's Proposition 1B.9 page 90. You can download the book from the author's web page: math.cornell.edu/~hatcher/AT/AT.pdf. $\endgroup$ – Philippe Gaucher Jun 27 '17 at 6:14
  • $\begingroup$ Ahh, yes, I see it now. I just had to read a little more carefully. $\endgroup$ – Bob Jun 27 '17 at 17:26
  • $\begingroup$ This result given in Philippe's comment has also a large generalisation discussed in Section 11.4 (and for the pointed case in Section 11.5) of the book advertised at groupoids.org.uk/nonab-a-t.html . (EMS 2011). $\endgroup$ – Ronnie Brown Jun 29 '17 at 20:22

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