True or False: If $F_1, F_2$ are fields and $f:F_1\to F_2$ is a ring homomorphism such that $f(1)=1$, then $f$ is injective.
I am not sure if this is true. Here's an attempt at a counter.
Consider $f:\mathbb{F}_4\to \mathbb{F}_2$ defined by $x\mapsto x^2$.
This map is well-defined since if $x-y=0\mod 2$, then $f(x)=f(y)$. It is a ring homomorphism since $f(x+y)=x^2+y^2 \mod 2$, and $f(xy)=(xy)^2=x^2y^2\mod 2$ since $F_2$ is commutative. It sends $1$ to $1$ since $1^2=1 \mod 2$. However it is not injective since the target is smaller than the domain.
Is this correct? I'm doubtful.
The statement is true: the kernel of $f$ must be a proper ideal of $F_1$ since $f(1)=1$, and the only proper ideal of $F_1$ is $(0)$.
The issue with your example is as follows: we can construct $\mathbb{F}_4$ by adjoining a root $\alpha$ of the polynomial $X^2+X+1$ to $\mathbb{F}_2$. Then $\alpha^2=\alpha+1\not\in\mathbb{F}_2$, so your homomorphism doesn't map into $\mathbb{F}_2$.