= Solution
Let $U$ be a four-dimensional <symplectic vector space>, with a <symplectic basis> $e_1,f_1,e_2,f_2$. Its invariant inverse-form bivector may be chosen as $\eta=e_1\wedge f_1+e_2\wedge f_2$. Put $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$, so $\eta\wedge\eta=2\mathrm{vol}$. Define a <symmetric bilinear form> on the six-dimensional <exterior square> by
$$
\xi\wedge\zeta=B(\xi,\zeta)\mathrm{vol}.
$$
It is symmetric because both factors have degree two. It is <nondegenerate>: in the six-element wedge <basis>, every $e_a\wedge e_b$ pairs nontrivially with the wedge of the complementary pair and with no other <basis> vector. Since $B(\eta,\eta)=2$, the <orthogonal complement>
$$
W=\eta^\perp\subset\Lambda^2 U
$$
has <dimension> five and inherits a <nondegenerate bilinear form>. It is the <primitive exterior square>, equivalently the <kernel> of the <symplectic contraction of an exterior square>.
The <symplectic Lie algebra> preserves $\eta$. It also preserves $\mathrm{vol}$, since its elements have <trace> zero and the induced action on $\Lambda^4U$ is multiplication by that <trace>. The <exterior-power Lie algebra representation> therefore preserves $B$ and $W$, giving a <Lie algebra homomorphism>
$$
\Phi:\mathfrak{sp}_4\longrightarrow\mathfrak{so}(W,B).
$$
To show injectivity, suppose $A$ acts as zero on $W$. It already acts as zero on the invariant line $\mathbb C\eta$, so it kills all of $\Lambda^2U=W\oplus\mathbb C\eta$. For any <basis> $u_1,\ldots,u_4$, this means
$$
Au_i\wedge u_j+u_i\wedge Au_j=0\qquad(i\ne j).
$$
If $k\ne i$, choose $j$ distinct from $i,k$; the coefficient of $u_k\wedge u_j$ forces $A_{ki}=0$. Thus $A$ is diagonal. The same identity gives $A_{ii}+A_{jj}=0$ for every pair. Applying this to three distinct indices forces each diagonal entry to vanish. Hence $A=0$ and $\Phi$ is injective.
Finally $\dim\mathfrak{sp}_4=2(2\cdot2+1)=10$ and $\dim\mathfrak{so}(W,B)=5\cdot4/2=10$. Injectivity between these equal-dimensional <vector spaces> gives surjectivity. Every <nondegenerate> complex <symmetric bilinear form> on a five-dimensional space is equivalent to the standard one. We have therefore constructed the <exceptional isomorphism between sp4 and so5>:
$$
\boxed{\mathfrak{sp}_4\cong\mathfrak{so}_5.}
$$
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