Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.