Question about maps on cofibers being zero

Question

I have the following problem, I am skeptical that the solution is as easy as the one I wrote so would be grateful if some expert can enlighten me.

Let $G\colon \mathcal{C}\to \mathcal{D}$ be a functor between stable $\infty$-categories (feel free to put more adjectives if needed) with left adjoint $F$, and let $\phi\colon X\to Y$ a map in $\mathcal{C}$ such that $G(\phi)\simeq 0$ in $\mathcal{D}$. Let $Z$ and $Z’$ be objects in $\mathcal{D}$ and consider a map $\psi\colon F(Z’)\to F(Z)$. Then we have a commutative diagram$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{l} \mathrm{map}_{\mathcal{C}}(F(Z),X) &\ra{} & \mathrm{map}_{\mathcal{C}}(F(Z'),X) \\ \qquad\da{\mathrm{map}(F(Z),\phi)} && \qquad\da{\mathrm{map}F(Z’),\phi)} \\ \mathrm{map}_{\mathcal{C}}(F(Z),Y) &\ra{}&\mathrm{map}_{\mathcal{C}}(F(Z’),Y) \\ \end{array} $$ By adjunction we have $\mathrm{map}(F(-),\phi)\simeq \mathrm{map}(-,G(\phi))$ therefore the two vertical maps above are homotopic to the zero map. Can I conclude that the map induced on the cofibers $$ \mathrm{map}(\mathrm{Fib}(\psi),\phi) \colon \mathrm{map}_{\mathcal{C}}(\mathrm{Fib}(\psi),X)\to \mathrm{map}_{\mathcal{C}}(\mathrm{Fib}(\psi),Y) $$ is also equivalent to the zero map? Notice that $\psi$ does not come from a map in $\mathcal{D}$.

Answer 1

The answer is no. Here is a counterexample. Let $p$ be a prime.

Let $F = \mathbb Z / p \otimes_{\mathbb Z}^L (-) : D(\mathbb Z) \to D(\mathbb Z)$, and let $\phi = p : \mathbb Z \to \mathbb Z$. Take $Z' = \mathbb Z, Z = \Sigma \mathbb Z$ to be appropriate shifts of $\mathbb Z$ and choose $\psi : \mathbb Z / p \to \Sigma \mathbb Z / p$ so that $Fib(\psi) = \mathbb Z / p^2$ (after all, there is a short exact sequence $0 \to \mathbb Z / p \to \mathbb Z / p^2 \to \mathbb Z / p \to 0$). Since $p = 0$ on $\mathbb Z / p$ -modules, but $p \neq 0$ on $\mathbb Z / p^2$ or its dual, you get a counterexample.