I'm trying to solve the problem $18 (a$ from the chapter $1$ of Hatcher's Algebraic topology. it is:
Prove that the wedge sum $S^1 \vee S^2$ has fundamental group $\mathbb{Z}$.
The next chapter is about Van Kampen,I don't know how to prove it without that.
Construct its universal cover. To this end let $X$ be the real line $\mathbb{R}$ with a copy $S^2_{(k)}$ of of the 2-sphere attached at each integer $k\in\mathbb{R}$.
The space $X$ is homotopy equivalent to the infinite wedge sum $\bigvee_{k\in\mathbb{Z}} S^2_{(k)}$, so is simply connected. The homotopy equivalence comes by just contracting each interval $[a,a+1]\subseteq\mathbb{R}$ to a point. We can see its simply connected because it has a cellular structure in which the 1-skeleton is a single point.
Now $\mathbb{Z}$ acts on $X$ as follows. For $n\in\mathbb{Z}$ and $t\in\mathbb{R}$ we let $t\cdot n=t+n$, and for $n\in\mathbb{Z}$ and $x_{(k)}\in S^2_{(k)}$ we let $x_{(k)}\cdot n=x_{(k+n)}\in S^2_{(k+n)}$. Observe that this is well defined, and moreover that it is free.
Now for the quotient we have a homeomorphism $X/\mathbb{Z}\cong S^1\vee S^2$. To see this, for $t\in\mathbb{R}$, we send the coset $t\cdot\mathbb{Z}$ to $exp(2\pi i t)\in S^1\hookrightarrow S^1\vee S^2$. For $x_{(k)}\in S^2_{(k)}$ we send the coset $x_{(k)}\cdot \mathbb{Z}$ to $x\in S^2\hookrightarrow S^1\vee S^2$. It should be clear that this is well defined and continuously invertible.
Hence we have a simply connected space $X$ with free right $\mathbb{Z}$ action, whose quotient is $S^1\vee S^2$. By the uniqueness of universal covers we conclude that $X$ is the universal cover of $S^1\vee S^2$, and moreover that $\pi_1(S^1\vee S^2)\cong \mathbb{Z}$.