Let $X$ and $Y$ be path connected and let $h : X → Y$ be a continuous function which induces the trivial homomorphism of fundamental groups. Let $x_0, x_1 ∈ X$ and let $f$ and $g$ be paths from $x_0$ to $x_1$. Prove that $h \circ f$ and $h \circ g$ are path homotopic.
My Attempt If $\gamma$ is any loop in $X$ based at $x_0$, $h_*[\gamma]=[e_{h(x_0)}]$ where $[e_{h(x_0)}]$ is the constant loop based at $h(x_0)$. In particular take two particular loops $f*\overline{g}$ and $f*\overline{f}$ (where $\overline{f}(t)=f(1-t)$ ) Now $$[e_{h(x_0)}]=h_*([g*\overline{f}])=[h\circ(g*\overline{f})]=[(h\circ g) * (h\circ \overline{f})]=[(h\circ g)] * [(h\circ \overline{f})] \ \ \ (*)$$ On the other hand $$[e_{h(x_0)}]=h_*([f*\overline{f}])=[h\circ(f*\overline{f})]=[(h\circ f) * (h\circ \overline{f})]=[(h\circ f)] * [(h\circ \overline{f})] \ \ \ (*)$$ Now can I compare $(*)$ and $(**)$ and conclude that $[(h\circ f)] = [(h\circ g)]$ ?
You can actually directly conclude with $(*)$: $$\begin{align} [h \circ g] * [h \circ \bar{f}] = [e_{h(x_0)}] & \implies [h \circ g] * \underbrace{[h \circ \bar{f}] * [h \circ f]}_{=[e_{h(x_0)}]} = [e_{h(x_0)}] * [h \circ f] \\ & \implies [h \circ g] = [h \circ f]. \end{align}$$