Embeddability of the cone of Klein bottle to $\mathbb R^4$

By using Alexander Duality, we can show that $K\not\hookrightarrow \mathbb R^3$ and we can also give an explicit formula for $K\hookrightarrow\mathbb R^4$ (here $K$ is the Klein bottle). How about the cone $CK$? Can it be embedded in $\mathbb R^4$?

It cannot be embedded in $\mathbb R^4$ 'nicely', which means if one considers any $\epsilon$-ball of the coning point, $K$ cannot be embedded into the boundary of the ball. I have also tried homology and cohomology exact sequences, and they did not give me any result.

The following lemma is an analogue of Corollary 3.46 in Hatcher’s “Algebraic Topology” and is proven by exactly the same method, using Alexander Duality (using cohomology with compact support instead of the ordinary cohomology as Hatcher does):

Lemma. Let $X$ be a locally contractible closed subset of $R^n$. Then $H_c^{n-2}(X)$ is torsion free.

Corollary. If $X$ is a nonorientable connected pseudomanifold of dimension $n-1$, it cannot be properly embedded in $R^n$.

Proof. Such $X$ will have $2$-torsion in $H_c^{n-2}(X,{\mathbb Z})$. qed

Now, turning to the open cone $CK$ over the Klein bottle, i.e. $(K\times [0,1))/(K\times 0)$. This space is nonorientable connected pseudomanifold. Each open subset of $CK$ containing the apex $a$ of this cone (the projection of $K\times 0$) is also a nonorientable pseudomanifold.

Suppose that $CK$ embeds in $R^4$. Take the intersection $Y$ of $CK$ with a sufficiently small open ball $B(a,r)$ in $R^4$. Use the homeomorphism $B(a,r)\to R^4$ to identify $Y$ with a closed subset $X\subset R^4$. We obtain a contradiction with Corollary above. Hence $CK$ does not embed in $R^4$.

Note that this argument is much more general applies to cones over compact nonorientable manifolds of any dimension.

Edit. Here is a better argument replacing Lemma and Corollary. I will use functoriality of the Alexander duality isomorphism AD: Given any closed subset $X\subset R^n$, we have a commutative square $$ \begin{array}{ccc} \check{H}^{n-1}_c(X; {\mathbb Z}) &\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}) \\ \alpha\downarrow & & \beta\downarrow \\ \check{H}^{n-1}_c(X; {\mathbb Z}/2)&\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}/2) \end{array} $$ Here cohomology is the compactly supported Chech cohomology. In the case when $X$ is locally contractible, it is naturally isomorphic to the compactly supported singular cohomology.

Now, if $X$ is homeomorphic to a nonorientable connected pseudomanifold of dimension $n-1$, then the homomorphism $\alpha$ is not surjective (since it is a homomorphism $0\to {\mathbb Z}/2$). On the other hand, the homomorphism $\beta$ is always surjective (as a homomorphism of $0$-degree reduced homology groups). Therefore, a nonorientable connected pseudomanifold of dimension $n-1$ cannot be properly topologically embedded in $R^n$. The rest of the argument is as above.