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Is there a left adjoint to the inclusion of the full subcategory of 1-connected spaces into the category of all spaces?

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A left adjoint would map the initial space, the empty space, to some initial simply connected space, but there is no initial simply connected space at all. In order to make your question more interesting, you should talk about pointed spaces in each case. – Martin Brandenburg Jul 14 '12 at 9:00
@MartinBrandenburg : Is the empty set not vacuously simply connected? – Joe Jul 14 '12 at 18:15
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Martin points out in the comments that one should include basepoints to make the question interesting. However, the inclusion functor $\text{1-connected based spaces} \to \text{based spaces}$ does not have a left adjoint because it does not preserve pullbacks. Consider the pullback in 1-connected based spaces of $f: (\mathbb R, 1) \rightarrow (\mathbb R, 0) \leftarrow (\mathbb R, 0) : g$ given by $f(x) = 1-x^2$ and $g(y) = y^2$. This pullback is some 1-connected space. But the pullback of the diagram in based spaces is $S^1$.

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It seems possible to me that the inclusion has a right adjoint, however. – Justin Young Jul 14 '12 at 12:15
It seems to me that you claim that the category of $1$-connected spaces has pullbacks. Can you comment why this should be true? Obviously they cannot be formed in spaces.. Thanks – mland Jul 15 '12 at 9:09
Fair point. The real problem here, though, is non-locally path connected, 1-connected spaces. – Justin Young Jul 15 '12 at 11:58

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