Number of homotopy classes of maps $X\to Y$ of CW-complexes Determine whether there are one, finitely many or infinitely many homotopy classes of maps $f\colon X\to Y$:

*

*$X=S^1\times S^1,\quad Y=S^3$

*$X=S^1\times S^1,\quad Y=S^1$

*$X=S^1\vee S^3,\quad Y=S^3$
My attempt
For (1) I viewed $X$ and $Y$ as absolute CW-complexes (where we get $X$ by glueing the boundary a 2-cell to the figure eight along $aba^{-1}b^{-1}$  and $Y$ by glueing a 3-cell to a point), so any cellular map $X\to Y$ must map $X$ to a point and since by cellular approximation every map is homotopy equivalent to a cellular map, we only have one class.
I'm not really sure how to approach (2). Maybe project to one of the components and then use that we can construct maps $S^1\to S^1$ of any degree? This argument doesn't really convince me.
For (3) my idea is similar. We can send $S^1$ to a point, and we know that there are infinitely many homotopy classes of maps $S^3\to S^3$, so this would imply that there are infinitely many homotopy classes of maps $S^1\vee S^3\to S^3$.
I'd appreciate any hints or comments on my solutions/attempts. Particularly for (2) and (3), I'd like to make my arguments rigorous.
 A: Your arguments for (1) and (3) are correct.
Also your approach to (2) is okay.
Let $p : S^1 \times S^1 \to S^1$ denote the projection onto the first factor and $i : S^1 \to S^1 \times S^1, i(z) = (z,*)$. Then $p\circ  i = id$.
There are infinitely many homotopy classes $\phi_n$ of maps $S^1 \to S^1$. This produces infinitely many homotopy classes $\phi_n \circ [p]$ of maps $S^1 \times S^1 \to S^1$: In fact, if  $\phi_n \circ [p] = \phi_m \circ [p]$, then $\phi_n  = \phi_n \circ [p] \circ  [i] = \phi_m \circ [p] \circ  [i] = \phi_m$, thus $n = m$.
A: Paul Frost's answer to (2) is much more straightforward, but here is another nice way to solve the second question, using Eilenberg-MacLane spaces:
Let's consider the set (which turns out to be a group) of based homotopy classes of maps $S^1 \times S^1 \to S^1$, denoted by $\langle S^1 \times S^1, S^1\rangle$. Then by theorem 4.57 in Algebraic Topology (Hatcher), we have an isomorphism
$$
\langle S^1 \times S^1, S^1 \rangle \cong H^1(S^1 \times S^1; \mathbb{Z}) \cong \mathbb{Z} \oplus \mathbb{Z},
$$
since $S^1$ is an Eilenberg-MacLane space $K(\mathbb{Z},1)$. Now, as sets, we clearly have an inclusion $\langle S^1 \times S^1, S^1 \rangle \subseteq [ S^1 \times S^1, S^1]$, meaning that $[ S^1 \times S^1, S^1]$ itself must be infinite.
