why is $H_0(A)\overset {i_*}{\to} H_0(X)$ injective? Let $X$ be a a topological space, $A\subset X$.
I've been told that it is "trivial" that if each path component of $X$ contains at most one path component of $A$ then $H_0(A)\overset {i_*}{\to} H_0(X)$ is injective.
But honestly I cannot see why it is true. tried to go with the definition, taking $a\in A$ such that $[a]_X=0$ that is we have $f:I\to X$ with $f(1)-f(0)=a$, and obiously $image(f)$ intersects only one path-component of $X$. and I am looking for a way to deduce that $image(f)$ is a subset of A and therefore $f\in C_1(A)$ so $[a]_A=0$.
Could you assist?

Let $a,b\in A\,.$ If $[a]_X=[b]_X\,,$ then $[a]_X-[b]_X$ is a boundary, hence there is a path in $X$ connecting $a$ to $b\,,$ so that $a,b$ lie in the same path connected component of $X\,.$ Since each path connected component of $X$ contains at most one path connected component of $A\,,$ $a,b$ can also be connected by a path in $A\,,$ hence we also have $[a]_A=[b]_A\,,$ and so the inclusion map is injective. 
