Every continuous $f:\mathbb{R}P^5\rightarrow (S^1\vee S^1)\times T^3$ is homotopic to a constant map. A practice exam question:

Show that every continuous map $f:\mathbb{R}P^5\rightarrow (S^1\vee S^1)\times T^3$ is homotopy equivalent to a constant map.

I'm not even sure where to start with this one. I've done similar questions before by lifting to a contractible covering space and then composing the contracting homotopy of $\text{Im}(f)$ to a point $\tilde{x}$ with the covering map $p$ to see that $f$ is also homotopic to a constant map that sends everything to $p(\tilde{x})=x$.
However, I'm not sure that strategy will work this time, as I'm not sure what the universal cover of $(S^1\vee S^1)\times T^3$ is. I know that a product of covering spaces is the covering space of a product, and I know the universal cover of $S^1\vee S^1$ is the Cayley Graph. I would guess that the universal cover of $T^3$ is $\mathbb{R}^3$, but I'm not sure.
So my guess for the universal cover of this thing is $\text{Cayley Graph} \times \mathbb{R}^3\simeq \text{Cayley Graph}$. But I don't think it's contractible.
Is there an alternative strategy I'm missing to tackle questions of this form? If not, is there a (reasonable) criterion for when the universal cover of a space is contractible?
Also does the domain of this map matter at all?
 A: I assume that $T^3=S^1\times S^1\times S^1$ is the $3$-torus.

I would guess that the universal cover of $T^3$ is $\mathbb{R}^3$

Correct.

So my guess for the universal cover of this thing is $\text{Cayley Graph} \times \mathbb{R}^3\simeq \text{Cayley Graph}$. But I don't think it's contractible.

It is. The Cayley graph of $\pi_1(S^1\vee S^1)\simeq F_2$ is a tree, and thus contractible just like every tree is.

If not, is there a (reasonable) criterion for when the universal cover of a space is contractible?

I doubt it. Every path connected, locally path connected and simply connected space is the universal covering of itself. Can we deduce anything about its contractibility? I don't think so.

Also does the domain of this map matter at all?

It matters, or rather what $f$ induces on the fundamental groups matters. Let $f:X\to Y$ be given with $X$ path connected and locally path connected. What you've missed is that given a covering $p:C\to Y$ the lifting exists if and only if $$f_\#(\pi_1(X))\subseteq p_\#(\pi_1(C))$$ If $C$ is universal then this means that the lifting exists if and only if $f_\#:\pi_1(X)\to\pi_1(Y)$ is the trivial map.
Note that it holds in this case, because $\pi_1(\mathbb{R}P^5)\simeq\mathbb{Z}/2\mathbb{Z}$ while $\pi_1((S^1\vee S^1)\times T^3)\simeq F_2\times\mathbb{Z}^3$ is torsion-free.
All in all: the standard approach works fine, although a bit more work is required.
