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Suppose $Y$ is an $(n-1)$-connected space, $n>2$, so we have Hurewicz isomorphisms $\pi_n(Y)\cong H_n(Y)$ and $\pi_{n-1}(\Omega Y)\cong H_{n-1}(\Omega Y)$. Let a map $\alpha\colon X\to\Omega Y$ be given. Naturally it induces a map $\beta\colon X\times S^1\to Y$. I want to show the following diagram is commutative: $$\require{AMScd} \begin{CD} H_{n-1}(X) @>\times[S^1]>> H_n(X\times S^1)\\ @V\alpha_*VV @V\beta_*VV \\ H_{n-1}(\Omega Y) @<\cong<< H_n(Y). \end{CD} $$

Since $\Omega Y$ is path connected, we may homotope $\alpha$ to a based map. Then $\beta$ factors though the reduced suspension $\Sigma X$. If $X=S^{n-1}$ is a sphere, the commutativity would then follow from tracking down the definition of $\pi_n(Y)\xrightarrow{\cong}\pi_{n-1}(\Omega Y)$. However I don't know how this helps for the general case.

One can also phrase the question in cohomology in the obvious way. (In particular the cross product $\times[S^1]$ will be replaced by the slant product $/[S^1]$.)

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The general case does follow from the case $Y=S^n$.

Without loss of generality $X=\Omega Y$ and $\alpha$ is the identity. Now the statement comes down to the assertion that the composition $$ \pi_{n-1}\Omega Y\to H_{n-1}\Omega Y\to H_n(\Omega Y\times S^1)\to H_n Y $$ corresponds to the Hurewicz map $\pi_nY\to H_nY$ via a natural isomorphism between $\pi_{n-1}\Omega Y$ and $\pi_nY$. This is true for all $Y$, and by naturality it is enough to prove it for $Y=S^n$.

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