multiplication in H-space and loopspace of the H-space

Question

Let $X$ be an $H-space$, and let a multiplication $,\cdot,$ be given, associative up to homotopy.

Let $\Omega X$ be the loopspace of $X$ based at the identity and let the multiplication $ \circ $ on the loopspace be given by concatenating loops.

We also have a multiplication on the loopspace given by pointwise multiplication of loops. Since the loops are based at the identity this also give a loop based at the identity. Call this multiplication $\mu$.

Are the multiplication maps $\mu$, $\circ$ homotopic?

I am trying to verify the assertion in Odd primary exponents of Moore Spaces that if the power map $\dot k: X \to X$ sending $x$ to a choice of the $k$th power $x^k=(x(x(x(x...))$ $k$ times, is null homotopic, then the map $k^\circ$ sending $\gamma \in \Omega X$ to its kth homotopy exponent is also null homotopic.

Define $k^\mu$ to be the $k$-th power map on the loopspace using pointwise multiplication of loops.

Now if $\mu$ and $\circ$ are indeed homotopic, then since $k^\mu$ is null homotopic, it follows that $k^\circ$ is also null homotopic.

So is it true that $\mu$ and $\circ$ are homotopic?

Answer 1

This follows by an Eckmann-Hilton argument and the Yoneda lemma. For any pointed space $Y$, $\mu$ and $\circ$ each give a binary operation on the set $[Y,\Omega X]$ of homotopy classes of pointed maps $Y\to\Omega X$. By Yoneda, $\mu$ and $\circ$ are homotopic iff these binary operations are equal for all $Y$.

Now observe that for any $Y$, the two binary operations on $[Y,\Omega X]$ both have the constant map as an identity, and they satisfy the interchange law $\mu(a\circ b,c\circ d)=\mu(a,c)\circ\mu(b,d)$ (since this equation holds on the nose for points of $\Omega X$, as can easily be verified by plugging in the definitions of $\mu$ and $\circ$). By Eckmann-Hilton, they must be equal. Explicitly, for any $a,b\in [Y,\Omega X]$ and $1\in[Y,\Omega X]$ the constant map, we have $$\mu(a,b)=\mu(a\circ 1,1\circ b)=\mu(a,1)\circ\mu(1,b)=a\circ b.$$

Answer 2

I too, was trying to answer this question, to understand a talk about homotopy exponents.

First the fact that the two multiplications are homotopic:

The equality $\mu(\gamma_1, \gamma_2) \circ \mu(\Gamma_1,\Gamma_2)=\mu((\gamma_1 \circ \Gamma_1),(\gamma_2 \circ\Gamma_2))$ holds

since both the LHS and RHS applied to $t \in [0,1/2]$ are $\gamma_1(t)\gamma_2(t).$

Both the LHS and RHS applied to $t \in [1/2,1]$ are $\Gamma_1(t) \Gamma_2(t).$

Although this equality is on-the-spot true, the two sided identity is only there for one of the multiplications up to homotopy. So we get that the two multiplications are homotopic.

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Here is what we can say about exponents on the h-space and the homotopy groups of the space: if $\circ^k$ is a nullhomotopic map then the kth power map in the fundamental group is 0.

More generally for all the homotopy groups we have that $\cdot: X\times X \to X$ induces $\tilde \mu: \pi_k(X)\times \pi_k(X) \cong \pi_k(X\times X) \to \pi_k(X)$. We have the higher homotopy group composition $\tilde \circ$ taking $a, b: S^n \to X$ to a $\tilde \circ b: S^n \to S^n \wedge S^n \xrightarrow{f \vee g} X$.

The eckman hilton argument is the same: the left hand side and right hand side are the same when applied both below and above the equator of the sphere.