In my algebraic topology course, we define a simply connected topological space as a topological space $X$ whose fundamental group $\pi_1(X)$ is reduces to the trivial group $\{1\}$ and my teacher told me that in such a space, if we consider two paths $\gamma_1:[0,1] \rightarrow X$, $\gamma_2:[0,1] \rightarrow X$ such that $\gamma_1(0) = \gamma_2(0) = x_0$ and $\gamma_1(1) = \gamma_2(1) = x$, then $\gamma_1, \gamma_2$ are homotopic.
I do not really understand why this is true. Of course, my first intuition was to concatenate these paths such that \begin{equation*} \gamma(t) = \begin{cases} \gamma_1(2t) &\text{for } t \in [0,1/2]\\ \gamma_2(2(1 - t)) & \text{for }t \in [1/2,1] \end{cases} \end{equation*} and by simply connectedness of $X$, we conclude that $\gamma$ is homotopic to the constant loop, but from here I do not see how to conclude that $\gamma_1$ and $\gamma_2$ are homotopic. Could someone give me a lead ? Thank you!
The property that is fundamental to show this result is that a space $X$ is simply connected if and only if for any continuous map $f : \mathbb{S}^1 \to X$, there exists a continuous map $F : D^2 \to X$ such that $F$ restricted to $\mathbb{S}^1 = \partial D^2$ is equal to $f$. It says that if you have a loop $L \in X$, you can find a topological disk in $X$ such that it is the boundary in $X$.
This can be used as a definition of a simply connected space.
Now, suppose $X$ is a simply connected space and suppose that $\gamma_0$ and $\gamma_1$ are paths with the same endpoints. Define $L$ as the concatenation (as in your question): it defines a map $f : \mathbb{S}^1 \to X$ that is continuous. Thus (recall $X$ is simply connected), there exists $F : D^2 \to X$ continuous that is equal to $f$ on the boundary.
Parametrizing the path $\gamma_0$ with $F(e^{i \pi t})$ and $\gamma_1$ with $F(e^{-i\pi t})$ for $t\in [0,1]$, define $\gamma_s(t)$ = $F((1-s)e^{i\pi t} + s e^{-i \pi t})$. This is an homotopy in $X$ between $\gamma_0$ and $\gamma_1$.