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Let us note $\textbf{Top}_{n}$ for the full subcategory of all $T_n$-spaces $n=0,...,4$. Construct left adoints to the inclusion functors $\textbf{Top}_{n+1}\rightarrow \textbf{Top}_n$ for $n=0,...,3$.

I can "solve" the cases $n=0,1,2$ using the adjoint functor theorem, but $\textbf{Top}_4$ is not as nice a category as the others (it does not have products, or, at least, the inclusion functor does not preserve them) and, therefore, I cannot use the adjoint fucntor theorem. Plus, the phrasing of the problem hints at an explicit construction of these left adjoints.

Hence, my question is how do I construct the left adjoints to these functor explicitly?

NB: This is problem V.9.4. from Mac Lane.

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    $\begingroup$ There's something odd here. If the inclusion functor does not preserve products, it cannot have a left adjoint. $\endgroup$ – Arnaud D. Jul 26 '17 at 18:41
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    $\begingroup$ I agree that this isn't possible. $\endgroup$ – Kevin Carlson Jul 26 '17 at 19:34
  • $\begingroup$ You are absolutely right! The question is then just simply wrong, which surprises me of a problem from Mac Lane. $\endgroup$ – JeCl Jul 26 '17 at 19:42
  • $\begingroup$ For the other ones, the idea is to simply take a quotient, identifying all points which aren't separated enough. You may have to iterate this process to stabilize at a sufficiently separated space. $\endgroup$ – Kevin Carlson Jul 26 '17 at 19:48
  • $\begingroup$ Is this CWM? Which page? $\endgroup$ – magma Jul 26 '17 at 21:19
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The inclusion $\mathbf{Top}_{4}\to\mathbf{Top}_3$ has no left adjoint, since $\mathbf{Top}_{4}$ is not closed under products and so cannot be a reflective subcategory of $\mathbf{Top}_{3}$.

All of the other left adjoints can be constructed by a transfinite induction where you step by step "throw out" any failure of your space to satisfy the required separation axiom. Let's start with the left adjoint to the inclusion $\mathbf{Top}_1\to\mathbf{Top}_0$. Given a space $X$ and two points $x,y\in X$ such that every open set containing $x$ contains $y$, any map from $X$ to a $T_1$ space must make $x$ and $y$ become equal. So take the equivalence relation generated by saying any two such points $x$ and $y$ are equivalent, and let $X_1$ be the quotient of $X$ by this equivalence relation. Every map from $X$ to a $T_1$ space factors uniquely through the quotient map $X\to X_1$. If $X_1$ is $T_1$, that means $X_1$ is the value of our desired left adjoint to $\mathbf{Top}_1\to\mathbf{Top}_0$ on $X$. If not, we can again take the quotient of $X_1$ by the equivalence relation that identifies all witnesses to its failure to be $T_1$. Call this quotient $X_2$.

We can repeat this process by transfinite induction to get a sequence of smaller and smaller quotients $X_\alpha$ of $X$, taking colimits at limit steps. This process must eventually terminate since $X$ is a set, so we eventually reach a quotient $X_\alpha$ which is $T_1$. Our left adjoint then sends $X$ to this space $X_\alpha$, since every map from $X$ to a $T_1$ space factors uniquely through the quotient map $X\to X_\alpha$.

For the left adjoint to $\mathbf{Top}_2\to\mathbf{Top}_1$, you do the exact same thing, except you identify pairs of points which witness the failure of the $T_2$ axiom instead of the $T_1$ axiom.

The left adjoint to $\mathbf{Top}_3\to\mathbf{Top}_2$ is more complicated, since it may not always take a space $X$ to a quotient. Given a space $X$ with topology $\tau$, let $\tau_1\subseteq\tau$ be the collection of open sets $U$ such that for each $x\in U$, there exist disjoint open sets $V$ and $W$ separating $x$ and $X\setminus U$. Note that if $f:X\to Y$ is any continuous map from $X$ to a $T_3$ space $Y$, then $f^{-1}(U)\in \tau_1$ for every open subset $U\subseteq Y$. Now let $\tau_2\subseteq \tau_1$ be the set of $U\in \tau_1$ such that for each $x\in U$, there exist disjoint $V,W\in \tau_1$ separating $x$ and $X\setminus U$. Again, if $f:X\to Y$ and $Y$ is $T_3$, then $f^{-1}(U)\in \tau_2$ for all open $U\subseteq Y$. We can iterate this by transfinite induction to get a descending sequence $\tau_\alpha$, taking intersections at limit steps. This eventually stabilizes, giving a topology $\tau_\alpha\subseteq\tau$ for that points and closed sets can be separated by open sets. Moreover, every $\tau$-continuous map to a $T_3$ space is also continuous with respect to $\tau_\alpha$.

The set $X$ with the topology $\tau_\alpha$ would then be the value of the left adjoint on $X$, except that points may not be closed in this topology. So we must now apply the construction described in the second and third paragraphs above to take the largest quotient of this topology which is $T_1$. This gives us a new space which we will call $X_1$.

Unfortunately, this space $X_1$ may still not be $T_3$, since taking the quotient may have destroyed the fact that points and closed sets can be separated by open sets. So we repeat the whole process over again with $X_1$, first shrinking the topology to separate points and closed sets, then taking a quotient to make points closed. This gives us a new space $X_2$. We repeat this whole process by transfinite induction, taking colimits at limit steps. At each step, we are taking a smaller quotient set of $X$ and/or a smaller topology, so it eventually stabilizes. Thus we reach a space $X_\alpha$ which is $T_3$, and every map from $X$ to a $T_3$ space factors uniquely through the canonical map $X\to X_\alpha$.

It is instructive to think about how this construction would fail to work if you tried to do it for the $T_4$ axiom instead of $T_3$. The problem is in defining $\tau_1$: a failure of the $T_4$ axiom is detected by a pair of open sets, and there's no way to decide which of the two should not be in $\tau_1$. You might think you can define $\tau_1$ to consist of those $U\in \tau$ such that for all $V\in\tau$ such that $U\cup V=X$, $X\setminus U$ and $X\setminus V$ can be separated by open sets. But this doesn't work, since a map $f:X\to Y$ to a $T_4$ space need not be continuous with respect to $\tau_1$. The problem is in saying "for all $V\in\tau$", since we really want to require this condition only for those $V$ which will be in our eventual $T_4$ topology, but we don't know what that topology is yet.

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  • $\begingroup$ What is your reference for the first sentence? I don't get this fact from the nlab page. e.g. $\mathbf{Top}_3$ does not have exponential objects. $\endgroup$ – Henno Brandsma Jul 27 '17 at 7:12
  • $\begingroup$ At the start of Section 4 of that page, it mentions that a reflective subcategory is always closed under limits that exist in the ambient category. $\endgroup$ – Eric Wofsey Jul 27 '17 at 7:17
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    $\begingroup$ As a quick proof of this fact, suppose the reflector $F$ exists and you have a diagram in the subcategory with a limit $L$ in the ambient category. Then $FL$ has the exact same universal property as $L$ in the ambient category (it corepresents cones over the diagram, since every such cone factors through the map $L\to FL$). So the map $L\to FL$ is an isomorphism, and so the object $L$ is in the subcategory (at least, up to isomorphism). $\endgroup$ – Eric Wofsey Jul 27 '17 at 7:20
  • $\begingroup$ OK, I see. So the $\mathbf{Top}_4$ embeddings are always going to be hard, as this category has awful limit properties. CWM only mentions $\mathbf{Top}_0$ being reflective in $\mathbf{Top}$, which is indeed true. $\endgroup$ – Henno Brandsma Jul 27 '17 at 7:31
  • $\begingroup$ It is exercise 5.9.4 in CWM, and it's refuted by this paper $\endgroup$ – Henno Brandsma Jul 27 '17 at 7:45
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The nlab page has some info and links under the reflection heading where constructions are discussed as well. But it seems, only for $n=0,1,2$. (Kolmogorov quotient, Hausdorff quotient etc.).

Maybe a similar construction to the Hausdorff quotient exists, identifying closed sets that can not be separated, somehow.

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