No. Here $T$ is the <pair of pants>, and part (b) gives $H_2(T)=0$, whereas
$$
H_2(S^1\times S^1)\cong H_2(T^2)\cong\mathbb Z.
$$
If $g:T^2\to T$ made $f\circ g$ <homotopic> to the <identity map> of the <torus>, <functoriality of homology> and <homotopy invariance of homology> would give
$$
(f\circ g)_*=f_*\circ g_*=(\operatorname{id}_{T^2})_*
$$
on $H_2(T^2)$. But $g_*$ factors through the zero group $H_2(T)$, so the left side is zero while the right side is the identity on $\mathbb Z$, a contradiction.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 21F / c / Solution

<Homotopic> maps induce the same homomorphism on every <homology group>. Consequently, <homotopy equivalent> spaces have isomorphic homology groups.

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 Incoming links: Homotopy invariance of homology