For $n \geq 2$, for $i = 1,\dots,n$, let $f_i \colon X_i \longrightarrow Y_i$ be maps of based $CW$-complexes, and consider a map $$f \colon \prod_{i=1}^n X_i \longrightarrow \prod_{i=1}^n Y_i$$ such that $f \simeq f_1 \times \dots \times f_n$.
Denote by $FW(Y_i) = \{(y_1,\dots,y_n) \; | \; y_i = * \text{ for at least one $i$} \}$, and by $\iota \colon FW(Y_i) \longrightarrow \prod_{i=1}^n Y_i$ the inclusion.
Suppose that there exists a map $\widehat{f} \colon \prod_{i=1}^n X_i \longrightarrow FW(Y_i)$ such that $f \simeq \iota \circ \widehat{f}$.
1.) Does this imply that for some $i$, $f_i \simeq *$?
2.) This seems like an "obvious" fact. I feel like there should be a very simple proof. But I can't see a simple mathematical reason why we shouldn't be able to "homotopy the product map into the subspace consisting of the fat-wedge". Is my intuition all wrong?
Put more succinctly, can a product of non-null-homotopic maps factor up to homotopy through the fat-wedge?
I do believe that I have a proof that the answer is "no", which does seem correct, but it's more involved than such a simple statement intuitively should need, and I'm almost believing that there must be a mistake simply because intuition says that it shouldn't be this complicated. To summarise it, my proof involves proving an equivalent statement about non-triviality of a composite $*X_i \longrightarrow FW(X_i) \longrightarrow FW(Y_i)$, which I do by calculating the homotopy cofibre, and proving that it is not homotopy equivalent to $\Sigma (\ast X_i) \vee FW(Y_i),$ which I do through suspending both spaces, when they each split as a wedge, and then inspecting homology.
For $n\geq2$ and distinct primes $p,q$ let $$X=S^n\cup_p e^{n+1},\qquad Y=S^n\cup_q e^{n+1},$$ where the cells are attached by maps of degrees $p,q$, respectively. The inclusion $$\iota:X\vee Y\hookrightarrow X\times Y$$ induces an isomorphism in homology. Since $X,Y$ are simply connected complexes $\iota$ is therefore a homotopy equivalence by the Whitehead Theorem.
Consider the maps $$i:S^n\hookrightarrow X,\qquad j:S^n\hookrightarrow Y$$ which include the bottom cells. The product mapping $$i\times j:S^n\times S^n\rightarrow X\times Y$$ deforms into $X\vee Y$. But both $i,j$ are essential.