Construction of an immersion of $T^3$ − point in $\mathbb{R}^3$? Let $p\in T^3$. How do I construct an immersion of $T^3\setminus\{p\}$ in $\mathbb{R}^3$?
 A: This is not terribly hard to see what's going on, though it's quite hard to write down a formula. Let's start by determining what $T^3 \setminus \{pt\}$ is.
First, look at its handlebody structure. This is a sort of thickened CW-structure. You start with a 3-ball $D^3$. Then you attach three "1-handles" $D^2 \times [0,1]$ by gluing on the 'sides' $D^2 \times \{0,1\}$ to the boundary of $D^3$; do this however you like, as long as the embeddings of $D^2$ are orientation-preserving. So far, we've just got a 3-ball with some handles on it. This is easy to embed in $\Bbb R^3$: whatever picture you've got is correct. Call this, I guess, $X^1$ so far.
Now one attaches three "2-handles" $D^1 \times D^2$ by the "boundary" $D^1 \times S^1$. Just like with the CW structure of the torus, these are attached by (thickenings) of the loops $[a,b]$, $[a,c]$, and $[b,c]$ in $\partial X^1 = \Sigma_3$. It's not too hard to make sure these loops are embedded in $\partial X^1$. Now let's attach the 2-handles to the embedding; this is where we're forced to make it an immersion. If you draw the picture, it's not terribly difficult to see how one might do this. (Sorry for the lack of details: It's hard to draw a picture of this, and I find it difficult to describe in words. The idea is to "contract" the loop $S^1$ to a point in a way that's an embedding of the disc $D^2$.)
Once you've done this three times, you have an immersion of $X^2$ into $\Bbb R^3$. What's left? Well, $\partial X^2 = S^2$, so to get $T^3$ you would attach a ball. This is impossible to do immersively. But to get $T^3 \setminus \{pt\}$ you just need to attach $S^2 \times [0,1)$! This is much easier. Suppose for convenience that our immersion never mapped to $0$. Then the immersion $i$ of $\partial X^2$ obviously extends to an immersion of $S^2 \times [0,1)$: just pick $i(x,t) = t \cdot i(x)$! 
One minor technicality to note is that when doing handlebodies, it looks like what you get at each stage is a manifold with corners. But you can just smooth those corners out as you go.
