# What are we allowed to do when proving deformation retraction of simplex?

I'm trying to show that the $$3$$-simplex with the edge identifications $$[v_0, v_1] \sim [v_2, v_3]$$ and $$[v_0, v_2] \sim [v_1, v_3]$$ deformation retracts onto the torus. I have a couple of potential solutions, but I'm not sure why/if they are correct.

The first one is inspired by this solution to a similar problem. This involves collapsing the face $$[v_0, v_1, v_2]$$ by identifying the edge $$[v_1, v_2]$$ with $$[v_1, v_0] + [v_0, v_2]$$, giving the "usual" scheme for a torus, as shown in the figure below. However, I'm not quite sure why this is allowed. What exactly happens with the edge $$[v_1, v_2]$$, shown in dark blue?

The other solution is to try to cut the $$3$$-simplex and then glue it together again in a way that shows that it is a torus, as inspired by this solution. I'm not really sure how to get this to work though. I always end up with weird things like shown in the figure below. What am I doing wrong?

In your first diagram, the right side of that diagram depicts the union of two faces $$[v_0,v_1,v_3] \cup [v_0,v_2,v_3] \subset \Delta^3$$ as a quadrilateral that I'll denote $$Q$$. You've also depicted the side identifications on $$Q$$, and I'm sure you can see that the quotient space $$Q /\!\!\sim$$ is homeomorphic to the torus. Good so far.
Furthermore, since $$Q$$ contains every point of $$\Delta^3$$ which is identified to some point other than itself, I'm sure you can see that there is an induced embedding $$Q /\!\!\sim \,\to \Delta^3 /\!\!\sim$$.
Putting these together, if you can find a strong deformation retraction from $$\Delta^3$$ to $$Q$$, i.e. a retraction map $$f : \Delta^3 \mapsto Q$$ and a homotopy from that map to the identity map on $$\Delta^3$$ such that points of $$Q$$ are stationary under that homotopy, then that map and that homotopy will respect all identifications, and therefore by passing to the quotient you will obtain a deformation retraction $$\Delta^3 /\!\!\sim \,\to Q /\!\!\sim$$.
So, can you find $$f$$? (Hint: There is a homeomorphism between $$\Delta^3$$ and the closed 3-ball that takes $$Q$$ to the lower hemisphere of $$\partial B^3 = S^2$$.)