Inclusion of path connected component induces isomorphism between fundamental groups.

Question

So, I am presented with an excercise that states: Prove that if $X_0 \subset X$ is the path-connected component of that contains $x_0$, then the inclusion $i: X_0 \to X$ induces an isomorphism between $\pi_1(X_0,x_0)$ and $\pi_1(X,x_0)$

My problem is that this seems pretty trivial, take $\phi([f])=[i\circ f]$, after checking it is well defined you can easily see that it is inyective.

And it is surjective because given $[f] \in \pi_1(X,x_0)$, you have that $f$ is a loop that starts in $x_0$ so its image must be in $X_0$ because you can connect all elements of $f([0,1])$ to $x_0$ by $f$. So defining $g:[0,1] \to X_0$ as $g(t)=f(t)\, \forall t\in [0,1]$. Then $i\circ g=f$ and $[i\circ g]=[f]$.

So, my quesion is, is this reasoning ok? I mean, a part from a few formalities this is quite trivial, so either it is that, or I am doing something wrong or missing something. Any help would be appreciated.

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your argument is correct.

Answer 2

As pointed out in a comment, it should be clarified that for injectivity

$[i \circ f] = [i \circ g]$ implies that we have a path homotopy $F: I^2 \rightarrow X$. Since $F(I^2) \subset Y$ by path-connectivity of $I^2$, $F$ restricted to $Y$ is continuous which gives the homotopy between $f,g$ thus $[f] = [g]$.