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Martin Brandenburg
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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.

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 $\alpha_{S^1}$.

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.

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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Martin Brandenburg
  • 63.1k
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  • 424

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 concretizablenot concretizable and hence has no separatorsmall 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 $\alpha_{S^1}$.

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.

What is the center of $\mathbf{hTop}$? I strongly believe that it is trivial, but it is hard to prove since $\mathbf{hTop}$ is not concretizable and hence has no 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 $\alpha_{S^1}$.

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

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 $\alpha_{S^1}$.

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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Martin Brandenburg
  • 63.1k
  • 12
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  • 424

The center of $\mathbf{hTop}$

What is the center of $\mathbf{hTop}$? I strongly believe that it is trivial, but it is hard to prove since $\mathbf{hTop}$ is not concretizable and hence has no 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 $\alpha_{S^1}$.

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