[proofplan] We must show that the formula $f_*([\gamma])=[f\circ\gamma]$ is meaningful on based homotopy classes, that it defines a group homomorphism, and that these homomorphisms respect identities and composition of based maps. Well-definedness follows by composing a based homotopy in $X$ with the map $f:X\to Y$. The homomorphism property follows because composition with $f$ commutes with concatenation of based loops. The functor laws then follow directly from identity maps and associativity of ordinary function composition. [/proofplan]

[step:Show that a based map sends based loops to based loops] Let $I:=[0,1]$. Let $f:(X,x_0)\to(Y,y_0)$ be a morphism in $\mathbf{Top}_*$, so $f:X\to Y$ is continuous and $f(x_0)=y_0$. Let \begin{align*} \gamma:I &\to X \end{align*} be a based loop at $x_0$, meaning $\gamma$ is continuous and $\gamma(0)=\gamma(1)=x_0$. The composite \begin{align*} f\circ\gamma:I &\to Y \end{align*} is continuous as a composite of continuous maps. Its endpoints satisfy \begin{align*} (f\circ\gamma)(0)=f(\gamma(0))=f(x_0)=y_0, \qquad (f\circ\gamma)(1)=f(\gamma(1))=f(x_0)=y_0. \end{align*} Thus $f\circ\gamma$ is a based loop at $y_0$. [/step]

[step:Prove that the induced map is well-defined on based homotopy classes] Define a set map \begin{align*} f_*:\pi_1(X,x_0)&\to\pi_1(Y,y_0)\\ [\gamma]&\mapsto [f\circ\gamma]. \end{align*} We prove that this definition is independent of the representative $\gamma$. Suppose $\gamma_0:I\to X$ and $\gamma_1:I\to X$ are based loops at $x_0$ with $[\gamma_0]=[\gamma_1]$ in $\pi_1(X,x_0)$. By definition, there exists a based homotopy \begin{align*} H:I\times I&\to X \end{align*} such that \begin{align*} H(s,0)&=\gamma_0(s),& H(s,1)&=\gamma_1(s),& H(0,t)&=x_0,& H(1,t)&=x_0 \end{align*} for all $s,t\in I$. Define \begin{align*} K:I\times I&\to Y\\ (s,t)&\mapsto f(H(s,t)). \end{align*} The map $K$ is continuous because $K=f\circ H$. For all $s,t\in I$, \begin{align*} K(s,0)&=f(H(s,0))=f(\gamma_0(s))=(f\circ\gamma_0)(s),\\ K(s,1)&=f(H(s,1))=f(\gamma_1(s))=(f\circ\gamma_1)(s),\\ K(0,t)&=f(H(0,t))=f(x_0)=y_0,\\ K(1,t)&=f(H(1,t))=f(x_0)=y_0. \end{align*} Hence $K$ is a based homotopy from $f\circ\gamma_0$ to $f\circ\gamma_1$. Therefore \begin{align*} [f\circ\gamma_0]=[f\circ\gamma_1] \end{align*} in $\pi_1(Y,y_0)$, so $f_*$ is well-defined. [/step]

[step:Verify that the induced map preserves loop concatenation] Let $\gamma:I\to X$ and $\delta:I\to X$ be based loops at $x_0$. Their concatenation \begin{align*} \gamma * \delta:I&\to X \end{align*} is defined by \begin{align*} (\gamma * \delta)(s)= \begin{cases} \gamma(2s), & 0\le s\le \frac{1}{2},\\ \delta(2s-1), & \frac{1}{2}\le s\le 1. \end{cases} \end{align*} For every $s\in I$, \begin{align*} (f\circ(\gamma * \delta))(s)= \begin{cases} f(\gamma(2s)), & 0\le s\le \frac{1}{2},\\ f(\delta(2s-1)), & \frac{1}{2}\le s\le 1, \end{cases} \end{align*} which is exactly the concatenation $(f\circ\gamma)*(f\circ\delta)$. Therefore \begin{align*} f\circ(\gamma * \delta)=(f\circ\gamma)*(f\circ\delta). \end{align*} Using the group operation in the fundamental group, we obtain \begin{align*} f_*([\gamma][\delta]) &=f_*([\gamma * \delta])\\ &=[f\circ(\gamma * \delta)]\\ &=[(f\circ\gamma)*(f\circ\delta)]\\ &=[f\circ\gamma][f\circ\delta]\\ &=f_*([\gamma])f_*([\delta]). \end{align*} Thus $f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)$ is a group homomorphism. [/step]

[step:Check that identity maps induce identity homomorphisms] Let $(X,x_0)$ be a pointed topological space, and let \begin{align*} \operatorname{id}_X:X&\to X \end{align*} be the identity map. This is a based map $(X,x_0)\to(X,x_0)$ because $\operatorname{id}_X(x_0)=x_0$. For every based loop $\gamma:I\to X$ at $x_0$, \begin{align*} (\operatorname{id}_X)_*([\gamma]) &=[\operatorname{id}_X\circ\gamma]\\ &=[\gamma]. \end{align*} Hence $(\operatorname{id}_X)_*=\operatorname{id}_{\pi_1(X,x_0)}$. [/step]

[step:Check that composition of based maps induces composition of homomorphisms] Let \begin{align*} f:(X,x_0)&\to(Y,y_0),& g:(Y,y_0)&\to(Z,z_0) \end{align*} be morphisms in $\mathbf{Top}_*$. Thus $f:X\to Y$ and $g:Y\to Z$ are continuous, $f(x_0)=y_0$, and $g(y_0)=z_0$. Their composite \begin{align*} g\circ f:X&\to Z \end{align*} is continuous and satisfies $(g\circ f)(x_0)=g(y_0)=z_0$, so $g\circ f:(X,x_0)\to(Z,z_0)$ is a based map. For every based loop $\gamma:I\to X$ at $x_0$, \begin{align*} (g\circ f)_*([\gamma]) &=[(g\circ f)\circ\gamma]\\ &=[g\circ(f\circ\gamma)]\\ &=g_*([f\circ\gamma])\\ &=(g_*\circ f_*)([\gamma]). \end{align*} Therefore \begin{align*} (g\circ f)_*=g_*\circ f_*. \end{align*} [/step]

[step:Conclude that the assignment is a functor] The preceding steps show that every object $(X,x_0)$ of $\mathbf{Top}_*$ is assigned a group $\pi_1(X,x_0)$, every morphism $f:(X,x_0)\to(Y,y_0)$ is assigned a group homomorphism $f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)$, identity morphisms are sent to identity homomorphisms, and composition of morphisms is preserved. Hence the assignment defines a covariant functor \begin{align*} \pi_1:\mathbf{Top}_*&\to\mathbf{Grp}. \end{align*} This is precisely the fundamental group functor. [/step]