Let $M$ be a topological monoid. $M$ can be considered as a category internal to topological spaces and has a simplicial space $N_\bullet(M)$ as its nerve. The geometric realization $BM=|N_\bullet(M)|$ of this simplicial space is the classifying space $BM$. The canonical map $\phi: M\to \Omega BM$ is constructed as follows (cf. Archipelago, canonical map of a monoid to its classifying space).
Construction: The standard n simplex $$\Delta^n=\{(x_0,...,x_n)\in\mathbb{R}^{n+1}|\sum_{i=0}^nx_i=1, 0\le x_i\le1\}$$ Let $m\in M$ and define $$l_m\colon \Delta^1\rightarrow BM$$ as the composition of $$\Delta^1\rightarrow M\times\Delta^1,(t_0,t_1)\mapsto(m,t_0,t_1)$$ the inclusion $$M\times\Delta^1\rightarrow \coprod\limits_{n\ge0}M^n\times\Delta^n$$ and the quotient map $$\coprod\limits_{n\ge0}M^n\times\Delta^n\rightarrow BM=(\coprod\limits_{n\ge0}M^n\times\Delta^n)/\tilde{}.$$ By precomposing with the homeomorphism $$[0,1]\rightarrow \Delta^1,t\mapsto(t,(1-t))$$ we get a path $$[0,1]\rightarrow BM$$ which I'll call $l_m$, too. Now we want to see that this is actually a loop: It holds $$l_m(0)=[m,(0,1)]=[e,0]=[m,(1,0)]=l_m(1)$$ where $[\_]$ denotes equivalence classes in $BM$ and the second and the third equality follows from the relations $\tilde{}$ we imposed on $BM$. It follows that $l_m$ factors as $$S^1=[0,1]/\{0,1\}\rightarrow BM,$$ which I'll call $l_m$, too. Define $$\phi: M\rightarrow \Omega BM, m\mapsto l_m.$$
My question: How to prove $\phi$ is a homomorphism of $H$-spaces (preserving the multiplication)? Then $\phi_*$ would be a homomorphism of pontrjagin rings.
Keep in mind the following picture:
This is supposed to represent a homotopy of paths $[0,1] \to \Delta^2$ between the path that goes $0 \to 1 \to 2$ and the path that goes straight $0 \to 2$. If you want a precise definition, let: $$\tilde{H}(s,t) = \begin{cases} (1-s) \cdot (1-2t, 2t, 0) + s \cdot (1-t, 0, t), & 0 \le t \le 1/2, \\ (1-s) \cdot (0, 2(1-t), 2t-1) + s \cdot (1-t, 0, t), & 1/2 \le t \le 1, \end{cases} \in \Delta^2.$$
As you can see, $\tilde{H}(0,t)$ is the concatenation of the straight paths $(1,0,0) \to (0,1,0)$ and $(0,1,0) \to (0,0,1)$ (in $\Delta^2$), whereas $\tilde{H}(1,t)$ is the straight path $(1,0,0) \to (0,0,1)$. It's also a based homotopy.
Now define: $$\begin{align} H : [0,1]^2 & \to M^2 \times \Delta^2 \to BM \\ (s,t) & \mapsto (m,m',\tilde{H}(s,t)) \end{align}$$ Then $H$ is a homotopy between the concatenation of $l_m \cdot l_{m'}$ and $l_{mm'}$ (you can check this on the formulas; use the simplicial relations of $BM$). In other words you have $l_{mm'} \sim l_m \cdot l_{m'}$, ie. $l$ is a map of H-spaces.