This follows by basic diagram chasing. If we have the following commuting triangle, with any two maps being equivalences, we can take pseudo-inverses to construct a pseudo-inverse to the third map.

Suppose we have the following commuting diagram where $g$ is an equivalence. We show $f$ is also an equivalence:

Since $g$ is an equivalence, there exists a functor $g'$ such that $gg' \cong 1_V$ and $g'g \cong 1_U$. The morphism $pg'j$ is then easily seen to be a pseudo-inverse to $f$, showing that it is also an equivalence.

Suppose the following diagram commutes and $g$ is an isofibration. We need to show that $f$ is an isofibration also:

Let $x \in X$ and $\alpha : fx \to y$ be an isomorphism in $Y$. We apply $j$ to get an isomorphism $j\alpha : jfx=gix \to jy$. As $g$ is an isofibration, there is an isomorphism $\beta : ix \to u \in U$ such that $g\beta = j\alpha$. Applying $p$, we have an isomorphism $p\beta : pix=x \to pu \in X$. If we apply $f$ to this map, we have: $$\begin{align*} fp\beta &= qg\beta && \text{right square}\\ &= qj\alpha && \text{definition of }\beta\\ &= \alpha. \end{align*}$$ Hence, given an isomorphism $\alpha$, there exists an isomorphism $p\beta$ such that $fp\beta = \alpha$, so $f$ is also an isofibration.

Suppose we have the following diagram in $\mathbf{Cat}$, where $g$ is a functor injective on objects. We need to show that $f$ is injective on objects also:

Suppose $x$ and $y$ are objects in $X$ such that $fx = fy$. We have the following chain of equalities: $$\begin{align*} fx = fy &\implies jfx=jfy\\ &\implies gix=giy && \text{(left square)}\\ &\implies ix=iy && (g \in C)\\ &\implies pix=piy\\ &\implies x=y. \end{align*}$$ Hence $f$ is also injective on objects.

Suppose that $F: \mathcal{A} \to \mathcal{B}$ is an isofibration and an equivalence. To see that it is surjective on objects, let $b \in B$. As $F$ is an equivalence, it is fully faithful, so there is some $a \in \mathcal{A}$ and isomorphism $\beta : Fa \to b \in \mathcal{B}$. As $F$ is an isofibration, there is an isomorphism $\alpha : a \to a' \in \mathcal{A}$ such that $F\alpha = \beta$. Then, we must have that $Fa' = b$, so $F$ is surjective on objects.

If $F$ is an equivalence that is surjective on objects, it is also an isofibration. Let $a \in \mathcal{A}$ and $\beta : Fa \to b$ be an isomorphism in $\mathcal{B}$. As $F$ is surjective on objects, $b = Fa'$ for some $a' \in \mathcal{A}$, so $\beta$ is a map $Fa \to Fa'$. By $F$ being full, there is a morphism $\alpha : a \to a'$ such that $F\alpha = \beta$. By $F$ being faithful, this $\alpha$ must be an isomorphism, so $F$ is an isofibration.

Suppose we have the following commuting diagram, where $f$ is injective on objects and $g$ is an equivalence and isofibration:

As this commutes, we have the following ``object square’’ commuting in $\mathbf{Set}$:

In particular, $f_0$ is injective and by Lemma 10, $g_0$ is surjective. As $(\text{inj}, \text{surj})$ form a weak factorisation system on $\mathbf{Set}$, there is a lift $h_0$. We now aim to turn $h_0$ into a functor. Let $\alpha: y \to y'$ be a morphism in $Y$. We then have a morphism $j\alpha : jy = gh_0y \to gh_0y'=jy' \in V_0$. As $g$ is fully faithful, there is a unique morphism $\beta : h_0 y \to h_0 y' \in U$ such that $g\beta = j\alpha$. We define $h\alpha := \beta$. As $j$ is a functor, it must preserve identity morphisms and as $g$ is fully faithful, the lift of an identity morphism from $v$ to $u$ must be the identity, so $h$ preserves identity morphisms. Similarly, by functorality of $j$ and fully faithfulness of $g$, we have that composites are mapped to composites, so $h$ is a functor and makes the bottom triangle commute:

The top triangle commutes on objects, so we need to check it commutes on maps. Given $\alpha : x \to x' \in X$, $$\begin{align*} g i \alpha &= j f \alpha\\ &= g h f \alpha \end{align*}$$ So $i \alpha = h f \alpha$ by fully faithfulness of $g$, meaning that the top triangle commutes. This means that $h$ is indeed a lift, which shows that functors injective on objects are in the left lifting class of fucntors that are equivalences and isofibrations.

Suppose we have the following diagram commuting, with $f$ an equivalence that is injective on objects and $g$ an isofibration:

To construct a functor $h:Y \to U$, we use the axiom of choice to get some extra structure. As $f$ is essentially surjective, for every object $y \in Y$, we choose an isomorphism $\alpha_y : fx_y \to y$, picking the identity morphism wherever possible, i.e $\alpha_{fx} = 1_{fx}$ (we call the argument of $f$ $x_y$, which is well defined at $f$ is injective on objects). As $g$ is an isofibration, we choose for every $y \in Y$ a morphism $\beta_y : ix_y \to u_y \in U$ such that $g\beta_y = j\alpha_y$, again, picking the identity morphism whenever possible, i.e $\beta_{fx} = 1_{ix}$. On objects, we define $h(y) = u_y$. For morphisms, the output is a little more complicated. Starting with a map $\gamma : y \to y'$, we form the composite $\alpha_{y'}^{-1} \gamma \alpha_y : fx_y \to fx_{y'}$. We let $\overline{\alpha_{y'}^{-1} \gamma \alpha_y} : x_y \to x_{y'}$ be the unique map that $f$ maps to $\alpha_{y'}^{-1} \gamma \alpha_y$. Then, applying $i$, we have a map $i \overline{\alpha_{y'}^{-1} \gamma \alpha_y} : ix_y \to ix_{y'}$. We can then compose with maps $\beta_y^{-1}$ and $\beta_{y'}$ to get $\beta_{y'} i \left(\overline{\alpha_{y'}^{-1} \gamma \alpha_y}\right) \beta_{y}^{-1} : u_y \to u_{y'}$. This is what we define $h\gamma$ to be.

To see that $h$ is a functor, we see that it sends the identity to the identity: $$\begin{align*} \beta_{y} i \left(\overline{\alpha_{y}^{-1} 1_y \alpha_y}\right) \beta_{y}^{-1} &= \beta_y i\left( \overline{1_y}\right) \beta_y^{-1} \\ &= \beta_y i(1_{x_y}) \beta_y^{-1}\\ &=1_{u_y}. \end{align*}$$ We also see that $h$ preserves composites: let $\gamma : y \to y'$ and $\delta : y' \to y''$. Then: $$\begin{align*} \beta_{y''} i \left(\overline{\alpha_{y''}^{-1} \delta \alpha_{y'}}\right) \beta_{y'}^{-1} \circ \beta_{y'} i \left(\overline{\alpha_{y'}^{-1}\gamma \alpha_y}\right) \beta_y^{-1} &= \beta_{y''} i \left( \overline{\alpha_{y''}^{-1} \delta \alpha_{y'}}\right) i \left( \overline{\alpha_{y'}^{-1}\gamma \alpha_y}\right) \beta_y^{-1}\\ &= \beta_{y''}i \left( \overline{\alpha_{y''}^{-1} \delta \alpha_{y'}} \circ \overline{\alpha_{y'}^{-1} \gamma \alpha_y} \right) \beta_y^{-1}\\ &= \beta_{y''} i \left( \alpha^{-1}_{y''} \delta \gamma \alpha_y \right) \beta_y^{-1} \end{align*}$$

We now show that it makes both the triangles commute:

Given $x \in X$, we need $ix = u_{fx}$, however this follows from our choice of maps being the identity wherever possible, which means $\beta_{fx} = 1_{ix}$. Likewise, for maps $\gamma: x \to x'$, our chosen isomorphisms are the identity, giving $hf\gamma = i\gamma$. Now, if $y \in Y$, we see by definition that $gu_y = jy$, so the bottom triangle commutes on objects. Given a map $\gamma : y \to y' \in Y$, by the properties of $\beta_y$ we have: $$\begin{align*} g\left( \beta_{y'} i \left(\overline{\alpha_{y'}^{-1} \gamma \alpha_y}\right) \beta_{y}^{-1} \right) &= g(\beta_{y'}) gi \left(\overline{\alpha_{y'}^{-1} \gamma \alpha_y}\right) g(\beta_{y})^{-1}\\ &= j(\alpha_{y'}) jf \left( \overline{\alpha_{y'}^{-1} \gamma \alpha_y} \right) j(\alpha_y)^{-1}\\ &= j(\alpha_{y'}) j(\alpha_{y'})^{-1} j\gamma j(\alpha_y) j(\alpha_y)^{-1}\\ &=j\gamma \end{align*}$$

Hence $h$ is a lift, as required.

Given $f:X \to Y$, we form a new category $Z$ with objects given by $X_0 \coprod Y_0$. We define the hom-sets as follows: $$\begin{align*} Z(x,y) := \begin{cases} x \in X, y \in X &\implies Y(fx, fy),\\ x \in X, y \in Y &\implies Y(fx, y),\\ x \in Y, y \in X &\implies Y(x, fy),\\ x \in Y, y \in Y &\implies Y(x, y). \end{cases} \end{align*}$$

We then have a functor $X \to Z$ which is the identity on objects, and $f$ on maps, which, in particular, is injective on objects. We can also construct another functor $Z \to Y$ which is formed by applying $f$ to objects from $X$ and the identity to objects from $y$. It is then the identity morphism on maps, which means it is fully faithful. It is also surjective on objects as every object in $Y$ is mapped to itself by this functor. The composite of these two morphisms is $f$ on objects and $f$ on maps, hence it is a factorisation of $f$ into a functor injective on objects, followed by a surjective equivalence (which by Lemma 10) is an equivalence and isofibration), as required.

Given $f:X \to Y$ we form the comma category $F \downarrow 1_Y$, then take the full subcategory spanned by objects $(x, y, \phi : fx \to y)$ where $\phi$ is an isomorphism in $Y$. Denote this category as $F \downarrow_{\cong} 1_Y$ Note that we have a functor $X \to F \downarrow_{\cong} 1_Y$ defined by sending $x \mapsto (x , fx, 1_{fx})$ and $\alpha : x \to x' \mapsto (\alpha, f\alpha)$. This is clearly injective on objects and fully faithful. We also have that every $(x, y, \phi) \cong (x, fx, 1_{fx})$ by the following commuting square in $Y$ (with horizontal maps being isomorphisms):

We have a functor $f \downarrow_{\cong} 1_Y \to Y$ given by projecting out the second component. This functor is also seen to be an isofibration: given $(x, y, \phi)$ and an isomorphism $\alpha : y \to y' \in Y$, the following square commutes, with horizontal maps isomorphisms, so is an isomorphism in $f \downarrow_{\cong} 1_Y$:

Our projection functor sends this isomorphism to $\alpha$, so it is an isofibration. The composite of the two functors defined is easily seen to be equal to $f$, completing the factorisation of $f$ as an equivalence that is injective on objects, followed by an isofibration.

Lemma 9 shows functors injective on objects are stable under retracts. Lemmas 9 and 9 show that equivalences that are isofibrations are also stable under retracts. Lemma 13 gives the desired lifting data and Lemma 14 gives the desired factorisation.

Lemmas 7 and 9 show that the class of equivalences that are injective on objects is stable under retracts. Lemma 8 shows the class of isofibrations are stable under retracts. Lemma 12 gives the desired lifting data and Lemma 14 gives the desired factorisation.

Lemma shows that $W$ satisfies 2-out-of-3. Corollaries 15 and 16 show that $(C \cap W, F)$ and $(C, F \cap W)$ are weak factorisation systems.