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What is the center of the homotopy category $\mathbf{hTop}$? I strongly believe that it is trivial, but it is hard to prove since $\mathbf{hTop}$ is not concretizable and hence has no small separator. This also means that currently I have no proof why the center is even a set ...

The center of $\mathbf{hTop}$ consists of homotopy classes of continuous maps $\alpha_X : X \to X$ for every space $X$. For every continuous map $f : X \to Y$ there should be a homotopy between $\alpha_Y \circ f$ and $f \circ \alpha_X$. These homotopies are not subject to any compatibility conditions when $f$ changes.

Clearly, $\alpha_X$ is (homotopic to) the identity when $X$ is contractible. Also, $\alpha_{\coprod_{i \in I} X_i}$ identifies with $\coprod_{i \in I} \alpha_{X_i}$, so the same holds when $X$ is a coproduct of contractible spaces. The easiest space which is not of this form is the circle $S^1$. I have no idea how to approach $\alpha_{S^1}$.

For $x \in X$, naturality with respect to $x : \star \to X$ shows that there is a path from $x$ to $\alpha_X(x)$.

Notice that for any functor $K : \mathbf{hTop} \to \mathcal{C}$ (for example $H_*,H^*,\pi_*$) we have a map $Z(\mathbf{hTop}) \to \mathrm{End}(K)$, so we get lots of operations.

I am open for $1$-categorical variations of the spaces, such as CW complexes, CGWH spaces or pointed spaces. But in this question I am not asking about higher categorical versions of the center.

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    $\begingroup$ Here's a nontrivial element of the center: For each space $X$, choose a point in each connected component, and let $\alpha_X: X\to X$ be the constant map to the chosen point on each connected component. The required diagrams do commute up to homotopy. $\endgroup$ Commented Nov 22, 2022 at 11:09
  • $\begingroup$ (This works in the homotopy category of CW complexes or Kan complexes, or equivalently in the weak homotopy category of all spaces, maybe not in the homotopy category of all spaces) $\endgroup$ Commented Nov 22, 2022 at 11:11
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    $\begingroup$ The weak homotopy category of all spaces is equivalent as a category to the homotopy category of CW complexes. You're right that the natural transformation I'm defining is not induced by something on the category of all topological spaces with continuous maps, but that doesn't matter, you can just work with CW complexes. (I'll admit that I was vage about what I mean by "space" in my comment) $\endgroup$ Commented Nov 22, 2022 at 13:05
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    $\begingroup$ Here's a better description: Note that in the homotopy category, a morphism $S\to X$ for a discrete space $S$ is the same as a map $S\to\pi_0(X)$ of sets. In other words, $\pi_0$ is also right adjoint to the functor taking a set to the discrete space. (The usual adjunction goes the other way). The counit of this adjunction gives a natural transformation $\pi_0(X)\to X$, with $\pi_0(X)$ viewed as discrete space. We also have the usual natural transformation $X\to\pi_0(X)$, compose those to get a natural map $X\to X$. $\endgroup$ Commented Nov 22, 2022 at 21:45
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    $\begingroup$ Are you still interested in the classical homotopy category? I can show in this case that $\alpha_{S^1}$ cannot be constant. Achim's analysis below then forces it to be the identity. I have no idea how to produce nontrivial elements of the centre in this case $\endgroup$
    – Tyrone
    Commented Nov 24, 2022 at 9:27

1 Answer 1

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Here's a partial answer constraining $\alpha_{S^1}$. First of all, I think my comment shows that (at least if we take $\operatorname{hTop}$ to be the homotopy category of CW complexes), it is possible for $\alpha_{S^1}$ to be constant. Let's think about what other degrees besides $0$ and $1$ are possible!

Assume $\alpha_{S^1}$ has degree $n$. By looking at the commutative diagrams for the quotient maps $(S^1)^{\times k}\to S^k$, we see that $\alpha_{S^k}$ has to be the degree $n^k$-map. Now look at the commutative diagram for $\eta: S^3\to S^2$. Precomposing $\eta$ by the degree $n^3$-map on $S^3$ yields the element $n^3\cdot \eta\in \pi_3(S^2)$. On the other hand, postcomposing $\eta$ with the degree $n^2$ map yields $n^4\cdot\eta\in\pi_3(S^2)$. So we learn $n^4=n^3$, and thus $n=0$ or $1$.

(That postcomposition on $\eta$ behaves in this quadratic way can be seen by viewing $\eta$ as the map $\mathbb{C^2}\setminus\{0\}\to \mathbb{C}P^1$ taking $(z,w)\mapsto \frac{z}{w}$, and realising that this commutes with taking $d$-th powers on either side. But on the right, $d$-th power is of degree $d$, on the left, of degree $d^2$.)

I'm not sure what to expect about the full description of the center. If $\alpha_{S^1}$ is the identity, we certainly see that all $\alpha_{S^k}$ are identities as above, but $\alpha$ could still do funny things on more complicated cell complexes. I think it does follow though that $\alpha$ induces identities on homology. Similarly, if $\alpha_{S^1}$ is the constant map, it follows that this holds on spheres and generally on homology, but maybe not that $\alpha_X$ is constant on each connected space (as in the example I gave in the comment).

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