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?

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.