Let $ (X,\mathscr O_X) $ and $ (Y,\mathscr O_Y) $ be ringed spaces over the same unspecified commutative ring.
My book defines a morphism between $ (X,\mathscr O_X) $ and $ (Y,\mathscr O_Y) $ as a pair $ (f,f^\flat) $ of a continuous function $ f\colon X\to Y $ and a morphism of sheaves $ f^\flat\colon \mathscr O_Y\to f_*\mathscr O_X $ (or, equivalently, a morphism $ f^\sharp\colon f^*\mathscr O_Y\to \mathscr O_X $). Composition of morphisms is defined as follows. Let
$$
(X,\mathscr O_X)\xrightarrow{(f,f^\flat)} (Y,\mathscr O_Y)\xrightarrow{(g,g^\flat)} (Z,\mathscr O_Z)
$$
be morphisms of ringed spaces. Then we define their composite $ (g,g^\flat)\circ (f,f^\flat) $ as the ringed space morphism $ (h,h^\flat)\colon (X,\mathscr O_X)\to (Z,\mathscr O_Z) $ where
$$
h = g\circ f\,\qquad h^\flat = g_*f^\flat\circ g^\flat
$$
as in the diagram
$$
\mathscr O_Z\xrightarrow{g^\flat} g_*\mathscr O_Y\xrightarrow{g_*f^\flat} g_*f_*\mathscr O_X
$$
where of course $ g_*f_*\mathscr O_X = (g\circ f)_*\mathscr O_X $.
Now, I was trying to show that a morphism $ (f,f^\flat)\colon (X,\mathscr O_X)\to (Y,\mathscr O_Y) $ is an isomorphism of ringed spaces iff $ f $ is homeo and $ f^\flat $ is an isomorphism of sheaves. This should be obvious, but my little brain is not working well enough right now.
What I tried to do
Let $ (g,g^\flat)\colon (Y,\mathscr O_Y)\to (X,\mathscr O_X) $ be an inverse of $ (f,f^\flat) $. Then
$$
\begin{aligned}
g\circ f &= 1_X\\
g_*f^\flat\circ g^\flat &= 1_{\mathscr O_X}
\end{aligned}\qquad
\begin{aligned}
f\circ g &= 1_Y\\
f_*g^\flat\circ f^\flat &= 1_{\mathscr O_Y}
\end{aligned}
$$
but this only tells me that $ f^\flat $ and $ g^\flat $ have a one sided inverse.
Best Answer
Given that $f_*g^\flat \circ f^\flat = 1_{\mathscr{O}_Y}$, the obvious candidate for the inverse of $f^\flat$ is $f_*g^\flat$. To get that $f^\flat \circ f_*g^\flat = 1_{f_*\mathscr{O}_X}$, it simply suffices to pushforward the morphism $g_*f^\flat \circ g^\flat = 1_{\mathscr{O}_X}$ by $f$: \begin{equation*} f^\flat \circ f_*g^\flat = (f \circ g)_*f^\flat \circ f_*g^\flat = f_*(g_*f^\flat \circ g^\flat) = f_*1_{\mathscr{O}_X} = 1_{f_*\mathscr{O}_X} \end{equation*}