I'm not sure this is research level so if this is not appropriate, feel free to move the question to StackExchange. However, I post it here since my "fake proof" is based on a (recent) paper and I'm not sure it would be fit for SE.
Everything is done $\infty$-categorically.
I'm using the paper A simple universal property of Thom ring spectra, by Antonin-Camarena and Barthel (see here for the arXiv version), and specifically using proposition 4.9 and corollary 5.9 of that paper.
For clarity, let me quote them here :
Prop 4.9 : if $f: S^{k+1}\to BGL_1(R)$ is a based map and $\overline f : \Omega^n\Sigma^n S^{k+1}\to BGL_1(R)$ is the corresponding $n$-fold loop map, then for any $A\in Alg_R^{\mathbb E_n}$, there is an equivalence of spaces $Map_{Alg_{R}^{\mathbb E_n}}(M\overline f,A) \simeq \Omega^{\infty +k+1}A$, if $A$ has characteristic $\chi(f)$, otherwise the space is empty.
Where $M\overline f$ denotes the Thom ring spectrum associated to $\overline f$, that is, the colimit of $\Omega^n\Sigma^nS^{k+1} \to BGL_1(R)\to Mod_R$
And
Corollary 5.9 : There is a map $\Omega^2\Sigma^2S^1\to BGL_1S^0$ whose associated Thom spectrum is equivalent to $H\mathbb Z$ as $\mathbb E_2$-ring spectra.
I changed the phrasing : in the paper it was written $\Omega^2S^3$, but I wrote it that way to emphasize the connection between the two results, and to emphasize the fact that a map is obtained as a $2$-fold loop map from $S^1$. So corollary 5.9 puts us in the situation of proposition 4.9 with $k+1=1$.
We get
Let $A\in Alg_\mathbb S^{\mathbb E_2}$, then if that space is nonempty, there is an equivalence of spaces $Map_{Alg_\mathbb S^{\mathbb E_2}}(H\mathbb Z,A) \simeq \Omega^{\infty +1}A$
My problem is the following : consider $K= H\mathbb Z\otimes_{\mathbb S}H\mathbb Z$. This is an $\mathbb E_\infty$-algebra (which is the coproduct of $H\mathbb Z$ with itself as $\mathbb E_\infty$-algebras), and in particular there are two inclusion map $H\mathbb Z\overset{in_i}\to K$, $i=0,1$ that are $\mathbb E_\infty$-maps.
Forgetting some structure, they are also $\mathbb E_2$-maps. But note that $Map_{Alg^{\mathbb E_2}}(H\mathbb Z, K) \simeq \Omega^{\infty +1}K$, so its $\pi_0$ is $\pi_1K = H_1(H\mathbb Z; \mathbb Z) = 0$ (indeed, one can easily prove that $H_{n+1}(K(\mathbb Z,n);\mathbb Z) = 0$ for all $n$, for instance by providing a cell-structure with no $n+1$-cells).
It follows that $Map_{Alg^{\mathbb E_2}}(H\mathbb Z, K)$ is connected, so $in_0\simeq in_1$ as $\mathbb E_2$-maps.
What this means is that $in_0\otimes in_1 : H\mathbb Z\otimes H\mathbb Z\to K\otimes K$ (tensor products will be over $\mathbb S$, unless explicitly stated) is equivalent to $i\otimes i$ as $\mathbb E_2$-maps, or even as maps of spectra, where $i : H\mathbb Z\to K$ is any $\mathbb E_2$-map (for instance $in_0$ or $in_1$)
But the identity of $K$ factors as $H\mathbb Z\otimes H\mathbb Z\overset{in_0\otimes in_1}\to K\otimes K\to K$ as a map of spectra, and so, as a map of spectra, it factors as $H\mathbb Z\otimes H\mathbb Z\overset{i\otimes i}\to K\otimes K\to K$.
But now this latter map can be upgraded to a map of $\mathbb E_\infty$-algebras : indeed if we chose $i=in_0$ for instance, we can see this latter map as $H\mathbb Z\otimes H\mathbb Z \to H\mathbb Z \overset{in_0}\to K$ where the first map is obtained from the coproduct property, applied to the identity two times.
This means that $id_K$ factors, as a map of spectra, through $H\mathbb Z$. Looking at homotopy groups, this is absurd (it would mean $\pi_iK = 0$ for $i\neq 0$, which is known to be false)
Of course there must be a mistake somewhere but I don't see where it is, so my question is :
Where is the mistake ?
