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

No. Here $T$ is the pair of pants, and part (b) gives $H_2(T)=0$, whereasIf $g:T^2\to T$ made $f\circ g$ homotopic to the identity map of the torus, functoriality of homology and homotopy invariance of homology would giveon $H_2(T^2)$. But $g_{\ast }$ 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.