Skip to main content
added 543 characters in body, clarifying content
Source Link
Jeremy Brazas
  • 7.2k
  • 1
  • 26
  • 47

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$Surjectivity is not likely to be surjective unless you impose more conditions on the spaces and/or map in questionmain difficulty but the overall approach is similar to that for covering maps. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). The difficulty is knowing when $\tilde{f}$ is continuous. The general answer is likely to be no in general but I can't think of an explicit counterexample at the moment.

If you are willing to allow $X$ to be simply connected and "$\Delta$-generated" (the quotient of a topological sum of simplices) then the answer is yes. To prove the continuity of $\tilde{f}$ you factor through the $\Delta$-generated coreflection of $\widetilde{Y}$. Thus if $X$ is a CW-complex, then you have an affirmative answer. If $X$ is simply connected, locally path-connected and first countable, then the answer is also affirmative.

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$ is not likely to be surjective unless you impose more conditions on the spaces and/or map in question. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). The difficulty is knowing when $\tilde{f}$ is continuous. The general answer is likely to be no but I can't think of an explicit counterexample at the moment.

If you are willing to allow $X$ to be simply connected and "$\Delta$-generated" (the quotient of a topological sum of simplices) then the answer is yes. To prove the continuity of $\tilde{f}$ you factor through the $\Delta$-generated coreflection of $\widetilde{Y}$. Thus if $X$ is a CW-complex, then you have an affirmative answer. If $X$ is simply connected, locally path-connected and first countable, then the answer is also affirmative.

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. Surjectivity is the main difficulty but the overall approach is similar to that for covering maps. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). The difficulty is knowing when $\tilde{f}$ is continuous. The answer is likely to be no in general but I can't think of an explicit counterexample at the moment.

If you are willing to allow $X$ to be simply connected and "$\Delta$-generated" (the quotient of a topological sum of simplices) then the answer is yes. To prove the continuity of $\tilde{f}$ you factor through the $\Delta$-generated coreflection of $\widetilde{Y}$. Thus if $X$ is a CW-complex, then you have an affirmative answer. If $X$ is simply connected, locally path-connected and first countable, then the answer is also affirmative.

added 543 characters in body
Source Link
Jeremy Brazas
  • 7.2k
  • 1
  • 26
  • 47

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$ is not likely to be surjective unless you impose more conditions on the spaces $Y$, $\widetilde{Y}$, or theand/or map $p$in question. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). I doubt thatThe difficulty is knowing when $\tilde{f}$ is continuous in. The general answer is likely to be no but I can't think of an explicit counterexample at the moment.

If you are willing to allow $X$ to be simply connected and "$\Delta$-generated" (the quotient of a topological sum of simplices) then the answer is yes. To prove the continuity of $\tilde{f}$ you factor through the $\Delta$-generated coreflection of $\widetilde{Y}$. Thus if $X$ is a CW-complex, then you have an affirmative answer. If $X$ is simply connected, locally path-connected and first countable, then the answer is also affirmative.

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$ is not likely to be surjective unless you impose more conditions on the spaces $Y$, $\widetilde{Y}$, or the map $p$. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). I doubt that $\tilde{f}$ is continuous in general.

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$ is not likely to be surjective unless you impose more conditions on the spaces and/or map in question. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). The difficulty is knowing when $\tilde{f}$ is continuous. The general answer is likely to be no but I can't think of an explicit counterexample at the moment.

If you are willing to allow $X$ to be simply connected and "$\Delta$-generated" (the quotient of a topological sum of simplices) then the answer is yes. To prove the continuity of $\tilde{f}$ you factor through the $\Delta$-generated coreflection of $\widetilde{Y}$. Thus if $X$ is a CW-complex, then you have an affirmative answer. If $X$ is simply connected, locally path-connected and first countable, then the answer is also affirmative.

Adding a little more detail.
Source Link
Jeremy Brazas
  • 7.2k
  • 1
  • 26
  • 47

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$ is not likely to be surjective unless you impose more conditions on the spaces $Y$, $\widetilde{Y}$, or the map $p$. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). I doubt that $\tilde{f}$ is continuous in general.

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$ is not likely to be surjective unless you impose more conditions on the spaces $Y$, $\widetilde{Y}$, or the map $p$.

Every fibration with totally path-disconnected fibers has the unique path lifting property (2.2.5 of E.H. Spanier, Algebraic Topology, McGraw-Hill, New York, 1966) implying that $p_{\ast}$ is injective. However, it seems to me that $p_{\ast}$ is not likely to be surjective unless you impose more conditions on the spaces $Y$, $\widetilde{Y}$, or the map $p$. In particular, for every based map $f:X\to Y$, there is a unique based function $\tilde{f}:X\to\widetilde{Y}$ such that $p\circ\tilde{f}=f$ (defined using the fact that every path has a unique lift). I doubt that $\tilde{f}$ is continuous in general.

Source Link
Jeremy Brazas
  • 7.2k
  • 1
  • 26
  • 47
Loading