What is the dimension of the suspension of $\mathbb{R}P^2$?

Question

The suspension $SX$ of a topological space $X$ is defined as follows: $${\displaystyle S(X)=(X\times I)/\{(x_{1},0)\sim (x_{2},0){\mbox{ and }}(x_{1},1)\sim (x_{2},1){\mbox{ for all }}x_{1},x_{2}\in X\}}.$$

My question is that: what is the dimension of $S(\mathbb{R}P^2)$, the suspension of real projective plane?

There are various ways to define dimension. Your question is not well-defined. — Арсений Кряжев, Jun 25 '22 at 06:02
@АрсенийКряжевiswithUkraine By dimension here I mean dimension of a polyhedron. — M.Ramana, Jun 25 '22 at 11:03

score 1 · Answer 1 · answered Jun 25 '22 at 05:53

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It is not correct to talk about the dimension of a suspension because it is not a manifold in our example. See this

answered Jun 25 '22 at 05:53

Mihail

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score 0 · Accepted Answer · answered Jun 25 '22 at 04:50

The dimension of the suspension is equal to the dimensions of $\mathbb{R}P^2$ plus $1$, so $3$. This is usually the case with 'nice' spaces - taking the suspension increases the dimension by $1$.

In fact, in terms of homology, taking the suspension "shifts" all of the (reduced) homology groups up by one dimension.

What is the dimension of the suspension of $\mathbb{R}P^2$?

2 Answers2