a subspace of $\mathbb R^3$ with $\pi_1=\mathbb Z_2$

I've been wondering about such problems.

1. It is well known that $\mathbb{RP}^2$ cannot be realized as a subspace of $\mathbb R^3$.

But does there exist a space $X\subset\mathbb R^3$ (maybe even $CW$-complex) with $\pi_1(X)=\mathbb Z_2$?

1. If it is possible, can we find such $X$ to be homotopy equivalent to $\mathbb{RP}^2$?

I think that the first questions seems to be not so hard to answer but my knowledge is not enough to answer it myself.

• Well, the long exact sequence of homology says that $0\to H_2(\mathbb{R}^3,X)\to H_1(X)\cong \mathbb{Z}_2\to 0$. – Neal Mar 16 '15 at 4:16
• So it just means that $H_2(\mathbb R^3,X)=\mathbb Z_2$ but I still cannot see the answer. Or $H_2(\mathbb R^3,X)=H_2(X)$ since $\mathbb R^3$ is contractible. Am I right? – pw1822 Mar 16 '15 at 19:00
• For $X$ compact, this is impossible by using Alexander duality; see mathoverflow.net/a/4482/33088 The other answers are also enlightening! – Takumi Murayama Mar 17 '15 at 3:02