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rgnrmllbrg
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Let $E$ be homotopy equivalent to a $k$-sphere. Let $q\colon E\to X$ be a map such that given any continuous $f\colon C\to X$ from a compact space $C$, there exists (a non-unique) $\tilde{f}\colon C\to E$ with $q\tilde{f}=f$. Assume also that $X$ is a connected CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.

Let $E$ be homotopy equivalent to a $k$-sphere. Let $q\colon E\to X$ be a map such that given any continuous $f\colon C\to X$ from a compact space $C$, there exists (a non-unique) $\tilde{f}\colon C\to E$ with $q\tilde{f}=f$. Assume also that $X$ is a CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.

Let $E$ be homotopy equivalent to a $k$-sphere. Let $q\colon E\to X$ be a map such that given any continuous $f\colon C\to X$ from a compact space $C$, there exists (a non-unique) $\tilde{f}\colon C\to E$ with $q\tilde{f}=f$. Assume also that $X$ is a connected CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.

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rgnrmllbrg
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Map from $\mathbb{S}^k$homotopy sphere with lifting property induces surjections on homotopy groups. Is it weak equivalence?

Let $q\colon \mathbb{S}^k\to X$$E$ be homotopy equivalent to a $k$-sphere. Let $q\colon E\to X$ be a map such that given any continuous $f\colon K\to X$$f\colon C\to X$ from a compact space $K$$C$, there exists (a non-unique) $\tilde{f}\colon K\to \mathbb{S}^k$$\tilde{f}\colon C\to E$ with $q\tilde{f}=f$. Assume also that $X$ is a CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.

Map from $\mathbb{S}^k$ with lifting property induces surjections on homotopy groups. Is it weak equivalence?

Let $q\colon \mathbb{S}^k\to X$ be a map such that given any continuous $f\colon K\to X$ from a compact space $K$, there exists (a non-unique) $\tilde{f}\colon K\to \mathbb{S}^k$ with $q\tilde{f}=f$. Assume also that $X$ is a CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.

Map from homotopy sphere with lifting property induces surjections on homotopy groups. Is it weak equivalence?

Let $E$ be homotopy equivalent to a $k$-sphere. Let $q\colon E\to X$ be a map such that given any continuous $f\colon C\to X$ from a compact space $C$, there exists (a non-unique) $\tilde{f}\colon C\to E$ with $q\tilde{f}=f$. Assume also that $X$ is a CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.

Source Link
rgnrmllbrg
  • 393
  • 1
  • 9

Map from $\mathbb{S}^k$ with lifting property induces surjections on homotopy groups. Is it weak equivalence?

Let $q\colon \mathbb{S}^k\to X$ be a map such that given any continuous $f\colon K\to X$ from a compact space $K$, there exists (a non-unique) $\tilde{f}\colon K\to \mathbb{S}^k$ with $q\tilde{f}=f$. Assume also that $X$ is a CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.