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?

Thanks in advance!
 A: 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 $\partial B^3 = S^2$.)
