At page 4 of Strom's "Modern Classical Homotopy Theory" there is a universal formulation of the algebraic closure of a field. You can read it here from google books.

Exercise 1.2a is then to convince yourself that the stated universal property does in fact characterize the algebraic closure of a field.

I cannot see why the $\bar f$ in the diagram should be unique. For example if $F=\mathbb R$ and $A=E=\mathbb C$, wouldn't the identity and complex conjugation both make the diagram commute?

Thanks in advance

P.S.: I haven't touch math in years, but always wanting to learn algebraic topology I decided to work through Strom's book for the fun of it; so I apologize if I'm missing something trivial. I tried to fish google for "universal property algebraic closure" but nothing came up.

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    $\begingroup$ You're completely correct! He messed up the definition here, and should have only asked for $\overline f$ to be unique up to a field automorphism of $A$ (fixing the image of $F$). $\endgroup$ – aes Nov 22 '14 at 3:20

The algebraic closure has no universal property at all (and therefore one should rather speak of an algebraic closure). Uniqueness of extended homomorphisms fails, and this is what Galois theory is all about. (I really wonder how such statements can find their way into an advanced book on homotopy theory published by the AMS.) By the way, Grothendieck's Galois theory offers a connection between Galois theory and covering theory. The corresponding statement in covering theory then is that the universal covering space of a given space actually has no universal property (and therefore, it should not be called universal).

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    $\begingroup$ (+1) Just to add, the problem is that in the category the book referenced the automorphism group of, say, $F\hookrightarrow \bar{F}$ is certainly not trivial (it's $\text{Gal}(\bar{F}/F)$!), and so $F\hookrightarrow\bar{F}$ can't be initial. $\endgroup$ – Alex Youcis Nov 22 '14 at 10:08
  • $\begingroup$ Yes, and basically Galois theory is about the group $\mathrm{Gal}(\overline{F}/F)$. $\endgroup$ – Martin Brandenburg Nov 22 '14 at 10:11
  • $\begingroup$ Yes, but if you consider pointed covering spaces, then in fact the universal covering space has two universal properties: it is initial in the category of pointed covering spaces and final in the category of simply-connected pointed spaces over the original space---see Section 4.18 of Bergmann's book on Universal Algebra (math.berkeley.edu/~gbergman/245). I wonder what the analogy is in Galois Theory? $\endgroup$ – Jonathan Gleason Jul 9 '17 at 22:23
  • $\begingroup$ @JonathanGleason: Uniqueness fails. $\endgroup$ – Martin Brandenburg Jul 14 '17 at 6:43

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