Let $\mathbf{Top}_*$ be the category of pointed topological spaces and $\mathbf{Top_1}$ the full subcategory of simply connected spaces. $\mathbf{Top}_*$ is complete and cocomplete. I am trying to show that $\mathbf{Top_1}$ is not complete.

If $X$ is simply connected but has a non-trivial $\pi_2$-group, then $\Omega X$ (the based loop space) is not simply connected. Also, $\Omega X$ is the pullback (in $\mathbf{Top}_*$) of the path space fibration $PX \to X$ along the map $* \to X$. For example, one could take $X = S^2$.

This shows that the pullback, if it exists in $\mathbf{Top_1}$, is not the same as the pullback computed in $\mathbf{Top}_*$.

Can I use this to show that the pullback doens't exist in $\mathbf{Top_1}$? Can I somehow show that if it did exist, then "it must be $\Omega X$"? Perhaps one can consider the relationship with the underlying sets, i.e. $\mathbf{Set}$?

  • $\begingroup$ It seems to me to be difficult to relate limits of simply connected pointed spaces with limits of their underlying sets. For one thing, the underlying set functor is not representable. $\endgroup$ – Zhen Lin Jun 23 '15 at 17:40
  • $\begingroup$ @ZhenLin I see, yes. Is there another way of going about this then? $\endgroup$ – user50948 Jun 23 '15 at 18:38
  • $\begingroup$ I've replaced $\mathbf{Top}$ (which almost always denotes the category of topological spaces) by $\mathbf{Top}_*$. $\endgroup$ – Martin Brandenburg Jun 23 '15 at 23:24
  • $\begingroup$ $\mathbf{Top}_1$ has products, right? Even infinite ones. So the only issue are pullbacks, or equalizers (these concepts are more or less equivalent if products exist). $\endgroup$ – Martin Brandenburg Jun 23 '15 at 23:31
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    $\begingroup$ @MartinBrandenburg $\mathbf{Top}_0$ is complete. The limit of a diagram in $\mathbf{Top}_0$ is the path component of the basepoint in the limit of the diagram in $\mathbf{Top}_*$. $\endgroup$ – Jeremy Rickard Jun 24 '15 at 12:25

Suppose a diagram of simply connected pointed spaces has limit $X$ in $\mathbf{Top}_*$. If it has limit $\tilde{X}$ in $\mathbf{Top}_1$ then there is a canonical map $\tilde{X}\to X$ that is the universal map from a simply connected pointed space to $X$.

The universal cover of $X$ (when it exists) comes close to this, but the failure of lifting theorems for spaces that are not locally path connected stops it working.

As noted in Jack Davies' comment, $X=S^1$ is the pullback of a pair of maps of simply-connected spaces. If a map $\tilde{X}\to X$ existed with the required universal property, then by considering paths (maps from $[0,1]$) and homotopies between them (maps from $[0,1]\times[0,1]$) it's fairly easy to see that it would have to be the universal cover $\mathbb{R}\to S^1$. However, there are maps $Y\to S^1$ from simply connected but not locally path connected spaces $Y$ that do not lift to maps $Y\to\mathbb{R}$.

  • $\begingroup$ thank you! I am a little confused by your remarks about how to show that the supposed map must be $\mathbb{R} \to S^1$. When you say "considering paths and homotopies..." is that in order to constructre a homeomorphism $\tilde X \to \mathbb{R}$? $\endgroup$ – user50948 Aug 24 '15 at 13:00
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    $\begingroup$ @user50948 Exactly. By the universal property we get a continuous map $\mathbb{R}\to\tilde{X}$. By the homotopy lifting property for the covering map $\mathbb{R}\to S^1$ we get an inverse (for a point $y\in\tilde{X}$ choose a path from the basepoint to $y$, map it into $S^1$ and lift to $\mathbb{R}$, this being well-defined since we can also lift homotopies between paths). The fact that this inverse $S^1\to\mathbb{R}$ is continuous follows from the fact a covering map is a local homeomorphism. $\endgroup$ – Jeremy Rickard Aug 24 '15 at 14:45

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