# Provide a three-fold connected non-regular covering of $\mathbb{S}^1 \vee \mathbb{T}^2$ along with its projection

Let $$\mathbb{S}^1 \vee \mathbb{T}^2$$ be the cicrle and the torus glued together at a single common point (wedge sum of the circle and the torus).

I am asked to provide a (connected) triple non-regular (i.e. not normal) covering of this space.

I know that the fundamental group of my space is the free product between the one of the circle and the one of the torus, i.e. $$\pi_1(\mathbb{S}^1 \vee \mathbb{T}^2) \simeq\pi_1(\mathbb{S}^1) *\pi_1(\mathbb{T}^2)=\mathbb{Z} *\mathbb{Z}^2$$ (immediate from Van-Kampen)

It isn't obvious to me which are the normal and non-normal subgroups, but I thought of the following drawing:

I heard that, roughly speaking, introducing some degree of asymmetry in my cover would increase its chances of being non-regular, so I decided to give special treatment to one point out of three in the fibre of my base point (which I chose to be the intersection point between the torus and the circle).

Here, the projection maps each "circle" (or rather, each third of helix) to the base circle and each "square" to the torus. Let's denote the generator of the circle $$\alpha$$, and the generators of the torus $$a$$ and $$b$$ (small and large loops respectively).

Is my drawing correct (i.e. does it represent a covering and is it non-regular)?

I can't see a reason why it should fail to be a covering. It might not be clear on the drawing, maybe I should have drawn the squares "sideways" so that the points in the fibre of the base point touch the torus at the center of the square rather than at $$a$$, because it looks like it is not in contact with $$b$$.

I think it is non-regular because I don't think it is possible to act transitively on the fibre, i.e., I don't think I can design a homeomorphism sending the second point in the fibre to the topmost one. After all, if I try to map the second square to the topmost square, I'll have a continuity problem when I reach the bottom square...

Can somebody confirm that this is the correct idea? If not, can somebody provide me with a three-fold non-regular cover of $$\mathbb{S}^1 \vee \mathbb{T}^2$$?