Perusing the computation of the homology of $\mathbb RP^2$ in J. Vick's Homology Theory, I am entangled with several doubts about its construction, as are listed below.

Doubt 1

The Mayer-Vietoris sequence is applied to compute its homology, and in the interior of $D^2$ an open cell $U$ is picked, together with a point $p$ contained in $U$. Set $V=\mathbb RP^2\setminus\{p\}$ and consider the Mayer-Vietoris sequence of the covering $\{U,V\}$. This book claims that

Both $U\cap V$ and $V$ have the homotopy type of $S^1$, wheras $U$ is contractible.

I am confused that how the author immediately asserts that $V$ has the homotopy type of $S^1$?

Doubt 2

The author continues the construction as follows

In the portion of the sequence given by $$H_1(U\cap V)\overset{\alpha}{\rightarrow}H_1(U)\oplus H_1(V)\overset{\beta}{\rightarrow}H_1(\mathbb RP^2)$$ it is easy to check that $\beta$ is an epimorphism. A generating one-cycle in $U\cap V$, when retracted out onto the boundary, is wrapped twice around $S^1$ since $f$ has degree two. Thus $\alpha$ is a monomorphism onto $2\mathbb Z$ and $H_1(\mathbb RP^2)\approx \mathbb Z_2$.

Here $f\colon S^1\to S^1$ is given by $f(z)=z^2$, and $\mathbb RP^2$ is identified with the identification space $D^2\cup_f S^1$. I don't quite understand the highlighted sentence, why "a generating one-cycle when retracted to the boundary is wrapped twice", and how does this fact imply that "$\alpha$ is onto $2\mathbb Z$"?

Doubt 3

The author then generalized the result to a proposition

If $f\colon S^{n-1}\to Y$ is continuous where $Y$ is Hausdorff, then there is an exact sequence $$\cdots\to H_m(S^{n-1})\to H_m(Y)\to H_m(Y_f)\to H_{m-1}(S^{n-1})\to\cdots\to H_0(S^{n-1})\to H_0(Y)\oplus\mathbb Z\to H_0(Y_f).$$

Here $Y_f=D^n\cup_f Y$ is the identification space obtained by attaching an $n$-cell $D^n$ to $Y$ via $f$. The author claims that

Away from the dimensions ($n$ and $n-1$), the addition of an $n$-cell does not affect the homology.

I am confused that if $n>1$, can it be inferred that $H_0(Y)\cong H_0(Y_f)$ from that exact sequence?

Any help is greatly appreciated.


2 Answers 2


$V$ is $S^2$ minus the north and south poles, with antipodal points identified. $S^2$ minus the north and south poles deformation retracts to the equator and this retraction commutes with the antipode operation. So $V$ deformation retracts to the equator (an $S^1$) with antipodal points identified (also an $S^1$).

One can interpret $U\cap V$ as the Arctic region minus the north pole, and the generator of $H_1(U\cap V)$ as some Arctic line of latitude, inside $U$ this homotopes into the equator, but the equator is twice the generator of $H_1(U)$ since in $U$ antipodal points on the sphere are identified.

For your last question, adding on a cell of dimension $n\ge2$ won't affect the number of path components of $Y$, so won't change the $H_0$.

  • $\begingroup$ Ah...Could you elaborate a little: 1. why retraction commutes with antipode operation? 2. $U$ is in the interior of $D^2$, are there antipodal points in $U$ on the sphere? Even so, why the equator is twice the generator of $H_1(U)$? (apologies for my ignorance) $\endgroup$
    – josephz
    Jan 17, 2018 at 17:46

For doubt 1: deleting one point from RP^{2} is same as deleting two points from S^{2}, which is homotopic to circle.


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