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Andreas Thom
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This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ deformation retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. I do not see how $\chi(X)=1$ could imply that $X$ is contractible. Indeed, any knot complement deformation retracts onto a 2-complex, which has the property that adding one 2-cell makes in contractible. It has Euler characteristic 1, but is not contractible.

This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ deformation retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. I do not see how $\chi(X)=1$ could imply that $X$ is contractible. Indeed, any knot complement deformation retracts onto a 2-complex, which has the property that adding one 2-cell makes in contractible. It has Euler characteristic 1, but is not contractible.

This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ deformation retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. I do not see how $\chi(X)=1$ could imply that $X$ is contractible.

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Andreas Thom
  • 25.5k
  • 4
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  • 142

This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ deformation retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. Now, if $\chi(X)=1$, then the first homology vanishes (at least rationally). I do not see how this$\chi(X)=1$ could imply that $X$ is contractible. Indeed, any knot complement deformation retracts onto a 2-complex, which has the property that adding one 2-cell makes in contractible. It has Euler characteristic 1, but is not contractible.

This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. Now, if $\chi(X)=1$, then the first homology vanishes (at least rationally). I do not see how this could imply that $X$ is contractible.

This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ deformation retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. I do not see how $\chi(X)=1$ could imply that $X$ is contractible. Indeed, any knot complement deformation retracts onto a 2-complex, which has the property that adding one 2-cell makes in contractible. It has Euler characteristic 1, but is not contractible.

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Andreas Thom
  • 25.5k
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  • 82
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This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. Now, if $\chi(X)=1$, then the first homology vanishes (at least rationally). I do not see how this could imply that $X$ is contractible.

This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

This answer concerns Question 1. Take a free group on $2$ generators and $k$ very long relations at random. Then, with high probability, the representation satisfies a small cancellation condition and the associated 2-complex $X$ is aspherical. Moreover, if this happens you also have $\chi(X) = k-1$.

A negative answer to Question 2 would also be a negative answer to Whitehead's Asphericity Conjecture. I believe that a positive answer is not easy to obtain either.

Concerning Question 3: If $X \subset Y$ are 2-complexes, then the induced map $H_2(X) \to H_2(Y)$ is injective. This follows from the long exact sequence in homology. If $Y$ retracts onto a 1-complex, then this implies that $H_2(X)=0$ and hence $\chi(X) \leq 1$ if $X$ is connected. Now, if $\chi(X)=1$, then the first homology vanishes (at least rationally). I do not see how this could imply that $X$ is contractible.

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Andreas Thom
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