Let $f: R_1 \rightarrow R_2$ be a ring homomorphism. Does $r \in R_1$ being non-invertible mean $f(r)$ is non-invertible?

Let $f: R_1 \rightarrow R_2$ be a ring homomorphism. Does $r \in R_1$ being non-invertible mean $f(r)$ is non-invertible?

This seems incorrect, but I find it difficult finding counter-example...

Thanks for any assistance!

Hint: Consider the inclusion $\Bbb Z \to \Bbb Q$.
Consider $f : \mathbb{R}\times (\mathbb{Z}/6\mathbb{Z}) \longrightarrow \mathbb{R}$, defined as $f: (a,b) \mapsto a$ (which is why we call it forgetful - it just forgets part of the structure).
Clearly, $\mathbb{R}$ is a field, so every element in $\mathbb{R}$ is invertible. But $(*,2)$ is not invertible in $\mathbb{R}\times \mathbb{Z}/6\mathbb{Z}$. In fact, every element in $\mathbb{R}$ is the image of a noninvertible element.
Hint: Start with the ring $\mathbb{Q}[x]$. of polynomials with rational coefficients. The polynomial $x+1$ is not invertible. You can find a nice homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}$ that sends $x+1$ to $1$.