Degree $1$ map from torus to sphere

I'm trying to find a smooth degree 1 map from the torus $T^2 = S^1 \times S^1$ to the $2$-sphere $S^2$.

My first thought was to use the two coordinates $(\theta_1,\theta_2)$ to map onto the usual spherical polar coordinates on the sphere - however I can't quite get that to work because it seems that one of these coordinates would cover the sphere twice and so give a degree $2$ map?

Alternatively I thought about using the winding number. So in general if we embed circles into $\mathbb{R}^3$ by the maps $f_1,f_2: S^1 \to \mathbb{R}^3$ we can then define a map $F:T^2 \to S^2$ by:

$$F(\theta_1,\theta_2) = \frac{f_1(\theta_1) - f_2(\theta_2)}{\| f_1(\theta_1) - f_2(\theta_2) \|}$$

However I'm now struggling to make suitable choices for $f_1$ and $f_2$, my guess is that I'd want to make the circles cross over each other once but I'm struggling to visualise this and write down a map so I would appreciate any help.

Let $(U, \varphi)$ be a chart on $T^2$, i.e. $U$ is an open subset of $T^2$ and $\varphi : U \to \mathbb{R}^2$ is a homeomorphism. Let $V$ be the open subset of $U$ such that $\varphi|_V : V \to B(0, 1)$ is a homeomorphism, i.e. $V = \varphi^{-1}(B(0, 1))$. Then $\varphi(\overline{V}) = \overline{\varphi(V)} = \overline{B(0, 1)}$ and $\varphi(\partial V) = \partial\varphi(V) = \partial B(0, 1) = S^1$.

Note that the quotient $\overline{B(0, 1)}/S^1$ is homeomorphic to $S^2$; let $\psi : \overline{B(0, 1)}/S^1 \to S^2$ be a homeomorphism. The composite $\psi\circ\varphi|_{\overline{V}} : \overline{V} \to S^2$ maps $\partial V$ to a single point, call it $p$, and $(\psi\circ\varphi|_{\overline{V}})|_V = \psi\circ\varphi|_V$ is a homeomorphism from $V$ to $S^2\setminus\{p\}$.

Now define $f : T^2 \to S^2$ by

$$f(x) = \begin{cases} \psi(\varphi(x)) & x \in \overline{V}\\ p & x \not\in \overline{V}. \end{cases}$$

Then $f$ is a continuous map. Furthermore, it has degree one .

More generally, we can use the same technique to construct a degree one map from any closed, connected, orientable $n$-manifold to $S^n$.

Here's a way to construct a degree one map.

Consider the standard CW complex for the torus consisting of a $0$-cell, two $1$-cells, and one $2$-cell. If we forget about the $2$-cell, what remains is $S^1\vee S^1$. That is $T^2 = (S^1\vee S^1)\cup_{\varphi} e^2$ where $e^2$ is the $2$-cell and $\varphi$ is the attaching map $\varphi : \partial e^2 \to S^1\vee S^1$.

Now consider the space $T^2/(S^1\vee S^1)$ where all the points of $S^1\vee S^1$ are identified with one another. This quotient space has the CW structure $e^0\cup_{\tilde{\varphi}} e^2$ where $e^0$ is a $0$-cell (i.e. a point) and $\varphi : \partial e^2 \to e^0$. We see that the quotient is $S^2$ and so we can view the quotient map as a map $f : T^2 \to S^2$.

As $f|_{T^2\setminus(S^1\vee S^1)} : T^2\setminus(S^1\vee S^1) \to S^2$ is a homeomorphism (it identifies the $2$-cell of $T^2$ with the $2$-cell of $S^2$), we see that $f$ is a degree one map $T^2 \to S^2$.

The above argument can be used to show that any connected closed smooth $n$-manifold $M$ admits a degree one map to $S^n$. Using Morse theory, one can show that $M$ admits a CW complex structure with only one $n$-cell, i.e. $M = M^{(n-1)}\cup_{\varphi}e^n$ where $M^{(n-1)}$ is the $(n-1)$-skeleton of $M$. Then $M \to M/M^{(n-1)} \cong S^n$ is a degree one map.

We needed to assume $M$ was smooth above in order to use Morse theory. However, for $n \neq 4$, every closed connected $n$-manifold has such a CW complex structure with one $n$-cell, even if it isn't smoothable, so all such manifolds admit a degree one map to $S^n$. If $n = 4$, it is not known whether such a CW complex structure exists (in fact, it is not even known if every closed four-manifold has a CW complex structure). See this MathOverflow question for more details.