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user43326
user43326
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If I am not mistaken, we get a counter-example to your last question easily.

First of all we have $\Omega Q\Sigma A=QA$ so we only need to find a map from $QA$ to $QB$ that is not a $H$-map to get a counter-example.

Denote by $i_X$ the standard map $X\rightarrow QX$ for spaces $X$. Then $i_QX :QX\rightarrow QQX$ is almost never a loop map. This can be seen by looking at the homology.

However, this doesn't give a counter example to your first question.

If I am not mistaken, we get a counter-example to your last question easily.

First of all we have $\Omega Q\Sigma A=QA$ so we only need to find a map from $QA$ to $QB$ that is not a $H$-map to get a counter-example.

Denote by $i_X$ the standard map $X\rightarrow QX$ for spaces $X$. Then $i_QX :QX\rightarrow QQX$ is almost never a loop map. This can be seen by looking at the homology.

If I am not mistaken, we get a counter-example to your last question easily.

First of all we have $\Omega Q\Sigma A=QA$ so we only need to find a map from $QA$ to $QB$ that is not a $H$-map to get a counter-example.

Denote by $i_X$ the standard map $X\rightarrow QX$ for spaces $X$. Then $i_QX :QX\rightarrow QQX$ is almost never a loop map. This can be seen by looking at the homology.

However, this doesn't give a counter example to your first question.

Source Link
user43326
user43326
  • 3.1k
  • 17
  • 25

If I am not mistaken, we get a counter-example to your last question easily.

First of all we have $\Omega Q\Sigma A=QA$ so we only need to find a map from $QA$ to $QB$ that is not a $H$-map to get a counter-example.

Denote by $i_X$ the standard map $X\rightarrow QX$ for spaces $X$. Then $i_QX :QX\rightarrow QQX$ is almost never a loop map. This can be seen by looking at the homology.