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I am not sure how to start the following question. Any help will be greatly appreciated! Thank you!!!

Let $K$ be a 3-dimensional simplicial complex, and let $f : K \rightarrow \mathbb{R}P^2 \times S^3$ and $g : \mathbb{R}P^2 \times S^3 \rightarrow S^1 \times  S^4$ be continuous maps. Show that the composite map $g \circ f$ is null homotopic.

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It's a bit difficult to answer without knowing what tools you have available. Here's the first approach that comes to mind; I apologize if it's overkill. First note that a map into a product is null-homotopic iff both components are, so you can deal separately with maps $K\to S^1$ and $K\to S^4$. The latter will be null-homotopic just because $K$ is only 3-dimensional (use simplicial approximation to move the image off of some point in $S^4$ and then deform away from that point). As for the map to $S^1$, I'd use that $S^1$ is the Eilenberg-Mac Lane space $K(\mathbb Z,1)$, i.e., a classifying space for the first cohomology functor with $\mathbb Z$ coefficients. But your map factors through $\mathbb RP^2\times S^3$, which that cohomology functor sends to zero.

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