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This answer has been edited so that it not only reads better but also gives a better answer. I'll refer to Dupont and Sah, "Scissors congruences, II" as [DS] and the paper of Sah and Wagoner, "Second cohomology of Lie groups made discrete" as [SW] in what follows.

Let $F$ be an algebraically closed field of characteristic zero; the question asks about $F=\mathbb{C}$ but the answer is true in at least this generality. Then $PGL(2,F)\cong PSL(2,F)$ sits in an exact sequence $$1 \to \mathbb{Z}/2 \to SL(2,F) \to PGL(2,F) \to 1.$$ So $K_2(F)=H_2(SL(2,F),\mathbb{Z})$ and $H_2(PGL(2,F),\mathbb{Z})\cong K_2(F)\oplus \mathbb{Z}/2$ see [DS (A5)] and [SW].

Let $G$ be the upper triangular matrices in $PGL(2,F)$ and let $U$ be the subgroup of $G$ consisting of upper triangular matrices with ones on the diagonal. So there is a split exact sequence $1 \to U \to G \to F^\times \to 1$, and $U$ is isomorphic to the additive group of $F$.

It is shown in [DS (A11)] that the inclusion $F^\times \to G$ and the quotient map $G \to F^\times$ induce an isomorphism $H_*(F^\times,\mathbb{Z}) \cong H_*(G,\mathbb{Z})$. In particular, it follows that $H_1(F^\times,U)=0$ (in your case, this is $H_1(\mathbb{C}^\times,\mathbb{C})=0$). Thus $H_*(G,\mathbb{Z})\cong \Lambda^*_{\mathbb{Z}}F^\times$, the exterior algebra on $F^\times$ sitting in degree one, plus a copy of $\mathbb{Q}/\mathbb{Z}$ in each odd degree from $3$ onwards coming from the Tor term in the Künneth theorem [DS, (A10)].

Now Matsumoto's theorem says that $K_2(F)$ is the quotient of $F^\times \otimes_{\mathbb{Z}}F^\times$ by the subgroup generated by elements of the form $a \otimes (1-a)$. These relations imply that $u\otimes v$ is equivalent to $-v\otimes u$, so $K_2(F)$ is really a quotient of $\Lambda^2_{\mathbb{Z}}F^\times$. So there is a surjective symmetrising map from $\Lambda^2_{\mathbb{Z}} F^\times$ to $K_2(F)$ sending $u\wedge v$ to the Steinberg symbol $\{u, v\}$ represented by $u\otimes v$. Looking at [DS (A2)], this is the map in the Bloch-Wigner theorem. In particular, it comes from the inclusion of the split torus into $SL(2,F)$, see [DS, top of p189], so this is your map.

The conclusion is that your map surjects onto $K_2(F)$, but that $H_2(PGL(2,F),\mathbb{Z})$ contains an extra factor of two. Since $\Lambda^2_{\mathbb{Z}}F^\times$ is a $\mathbb{Q}$-vector space, I don't think the extra factor of two can be hit by $H_2(G,\mathbb{Z})$, but I admit that this confuses me.

This answer has been edited so that it not only reads better but also gives a better answer. I'll refer to Dupont and Sah, "Scissors congruences, II" as [DS] and the paper of Sah and Wagoner, "Second cohomology of Lie groups made discrete" as [SW] in what follows.

Let $F$ be an algebraically closed field of characteristic zero; the question asks about $F=\mathbb{C}$ but the answer is true in at least this generality. Then $PGL(2,F)\cong PSL(2,F)$ sits in an exact sequence $$1 \to \mathbb{Z}/2 \to SL(2,F) \to PGL(2,F) \to 1.$$ So $K_2(F)=H_2(SL(2,F),\mathbb{Z})$ and $H_2(PGL(2,F),\mathbb{Z})\cong K_2(F)\oplus \mathbb{Z}/2$ see [DS (A5)] and [SW].

Let $G$ be the upper triangular matrices in $PGL(2,F)$ and let $U$ be the subgroup of $G$ consisting of upper triangular matrices with ones on the diagonal. So there is a split exact sequence $1 \to U \to G \to F^\times \to 1$, and $U$ is isomorphic to the additive group of $F$.

It is shown in [DS (A11)] that the inclusion $F^\times \to G$ and the quotient map $G \to F^\times$ induce an isomorphism $H_*(F^\times,\mathbb{Z}) \cong H_*(G,\mathbb{Z})$. In particular, it follows that $H_1(F^\times,U)=0$ (in your case, this is $H_1(\mathbb{C}^\times,\mathbb{C})=0$). Thus $H_*(G,\mathbb{Z})\cong \Lambda^*_{\mathbb{Z}}F^\times$, the exterior algebra on $F^\times$ sitting in degree one, plus a copy of $\mathbb{Q}/\mathbb{Z}$ in each odd degree from $3$ onwards coming from the roots of unity [DS, (A10)].

Now Matsumoto's theorem says that $K_2(F)$ is the quotient of $F^\times \otimes_{\mathbb{Z}}F^\times$ by the subgroup generated by elements of the form $a \otimes (1-a)$. These relations imply that $u\otimes v$ is equivalent to $-v\otimes u$, so $K_2(F)$ is really a quotient of $\Lambda^2_{\mathbb{Z}}F^\times$. So there is a surjective symmetrising map from $\Lambda^2_{\mathbb{Z}} F^\times$ to $K_2(F)$ sending $u\wedge v$ to the Steinberg symbol $\{u, v\}$ represented by $u\otimes v$. Looking at [DS (A2)], this is the map in the Bloch-Wigner theorem. In particular, it comes from the inclusion of the split torus into $SL(2,F)$, see [DS, top of p189], so this is your map.

The conclusion is that your map surjects onto $K_2(F)$, but that $H_2(PGL(2,F),\mathbb{Z})$ contains an extra factor of two. Since $\Lambda^2_{\mathbb{Z}}F^\times$ is a $\mathbb{Q}$-vector space, I don't think the extra factor of two can be hit by $H_2(G,\mathbb{Z})$, but I admit that this confuses me.

This answer has been edited so that it not only reads better but also gives a better answer. I'll refer to Dupont and Sah, "Scissors congruences, II" as [DS] and the paper of Sah and Wagoner, "Second cohomology of Lie groups made discrete" as [SW] in what follows.

Let $F$ be an algebraically closed field of characteristic zero; the question asks about $F=\mathbb{C}$ but the answer is true in at least this generality. Then $PGL(2,F)\cong PSL(2,F)$ sits in an exact sequence $$1 \to \mathbb{Z}/2 \to SL(2,F) \to PGL(2,F) \to 1.$$ So $K_2(F)=H_2(SL(2,F),\mathbb{Z})$ and $H_2(PGL(2,F),\mathbb{Z})\cong K_2(F)\oplus \mathbb{Z}/2$ see [DS (A5)] and [SW].

Let $G$ be the upper triangular matrices in $PGL(2,F)$ and let $U$ be the subgroup of $G$ consisting of upper triangular matrices with ones on the diagonal. So there is a split exact sequence $1 \to U \to G \to F^\times \to 1$, and $U$ is isomorphic to the additive group of $F$.

It is shown in [DS (A11)] that the inclusion $F^\times \to G$ and the quotient map $G \to F^\times$ induce an isomorphism $H_*(F^\times,\mathbb{Z}) \cong H_*(G,\mathbb{Z})$. In particular, it follows that $H_1(F^\times,U)=0$ (in your case, this is $H_1(\mathbb{C}^\times,\mathbb{C})=0$). Thus $H_*(G,\mathbb{Z})\cong \Lambda^*_{\mathbb{Z}}F^\times$, the exterior algebra on $F^\times$ sitting in degree one.

Now Matsumoto's theorem says that $K_2(F)$ is the quotient of $F^\times \otimes_{\mathbb{Z}}F^\times$ by the subgroup generated by elements of the form $a \otimes (1-a)$. These relations imply that $u\otimes v$ is equivalent to $-v\otimes u$, so $K_2(F)$ is really a quotient of $\Lambda^2_{\mathbb{Z}}F^\times$. So there is a surjective symmetrising map from $\Lambda^2_{\mathbb{Z}} F^\times$ to $K_2(F)$ sending $u\wedge v$ to the Steinberg symbol $\{u, v\}$ represented by $u\otimes v$. Looking at [DS (A2)], this is the map in the Bloch-Wigner theorem. In particular, it comes from the inclusion of the split torus into $SL(2,F)$, see [DS, top of p189], so this is your map.

The conclusion is that your map surjects onto $K_2(F)$, but that $H_2(PGL(2,F),\mathbb{Z})$ contains an extra factor of two. Since $\Lambda^2_{\mathbb{Z}}F^\times$ is a $\mathbb{Q}$-vector space, I don't think the extra factor of two can be hit by $H_2(G,\mathbb{Z})$, but I admit that this confuses me.

This answer has been edited so that it not only reads better but also gives a better answer. I'll refer to Dupont and Sah, "Scissors congruences, II" as [DS] and the paper of Sah and Wagoner, "Second cohomology of Lie groups made discrete" as [SW] in what follows.

Let $F$ be an algebraically closed field of characteristic zero; the question asks about $F=\mathbb{C}$ but the answer is true in at least this generality. Then $PGL(2,F)\cong PSL(2,F)$ sits in an exact sequence $$1 \to \mathbb{Z}/2 \to SL(2,F) \to PGL(2,F) \to 1.$$ So $K_2(F)=H_2(SL(2,F),\mathbb{Z})$ and $H_2(PGL(2,F),\mathbb{Z})\cong K_2(F)\oplus \mathbb{Z}/2$ see [DS (A5)] and [SW].

Let $G$ be the upper triangular matrices in $PGL(2,F)$ and let $U$ be the subgroup of $G$ consisting of upper triangular matrices with ones on the diagonal. So there is a split exact sequence $1 \to U \to G \to F^\times \to 1$, and $U$ is isomorphic to the additive group of $F$.

It is shown in [DS (A11)] that the inclusion $F^\times \to G$ and the quotient map $G \to F^\times$ induce an isomorphism $H_*(F^\times,\mathbb{Z}) \cong H_*(G,\mathbb{Z})$. In particular, it follows that $H_1(F^\times,U)=0$ (in your case, this is $H_1(\mathbb{C}^\times,\mathbb{C})=0$). Thus $H_*(G,\mathbb{Z})\cong \Lambda^*_{\mathbb{Z}}F^\times$, the exterior algebra on $F^\times$ sitting in degree one, plus a copy of $\mathbb{Q}/\mathbb{Z}$ in each odd degree from $3$ onwards coming from the Tor term in the Künneth theorem [DS, (A10)].

Now Matsumoto's theorem says that $K_2(F)$ is the quotient of $F^\times \otimes_{\mathbb{Z}}F^\times$ by the subgroup generated by elements of the form $a \otimes (1-a)$. These relations imply that $u\otimes v$ is equivalent to $-v\otimes u$, so $K_2(F)$ is really a quotient of $\Lambda^2_{\mathbb{Z}}F^\times$. So there is a surjective symmetrising map from $\Lambda^2_{\mathbb{Z}} F^\times$ to $K_2(F)$ sending $u\wedge v$ to the Steinberg symbol $\{u, v\}$ represented by $u\otimes v$. Looking at [DS (A2)], this is the map in the Bloch-Wigner theorem. In particular, it comes from the inclusion of the split torus into $SL(2,F)$, see [DS, top of p189], so this is your map.

The conclusion is that your map surjects onto $K_2(F)$, but that $H_2(PGL(2,F),\mathbb{Z})$ contains an extra factor of two. Since $\Lambda^2_{\mathbb{Z}}F^\times$ is a $\mathbb{Q}$-vector space, I don't think the extra factor of two can be hit by $H_2(G,\mathbb{Z})$, but I admit that this confuses me.

This answer has been edited so that it not only reads better but also gives a better answer. I'll refer to Dupont and Sah, "Scissors congruences, II" as [DS] and the paper of Sah and Wagoner, "Second cohomology of Lie groups made discrete" as [SW] in what follows.

Let $F$ be an algebraically closed field of characteristic zero; the question asks about $F=\mathbb{C}$ but the answer is true in at least this generality. Then $PGL(2,F)\cong PSL(2,F)$ sits in an exact sequence $$1 \to \mathbb{Z}/2 \to SL(2,F) \to PGL(2,F) \to 1.$$ So $K_2(F)=H_2(SL(2,F),\mathbb{Z})$ and $H_2(PGL(2,F),\mathbb{Z})\cong K_2(F)\oplus \mathbb{Z}/2$ see [DS (A5)] and [SW].

Let $G$ be the upper triangular matrices in $PGL(2,F)$ and let $U$ be the subgroup of $G$ consisting of upper triangular matrices with ones on the diagonal. So there is a split exact sequence $1 \to U \to G \to F^\times \to 1$, and $U$ is isomorphic to the additive group of $F$.

It is shown in [DS (A11)] that the inclusion $F^\times \to G$ and the quotient map $G \to F^\times$ induce an isomorphism $H_*(F^\times,\mathbb{Z}) \cong H_*(G,\mathbb{Z})$. In particular, it follows that $H_1(F^\times,U)=0$ (in your case, this is $H_1(\mathbb{C}^\times,\mathbb{C})=0$). Thus $H_*(G,\mathbb{Z})\cong \Lambda^*_{\mathbb{Z}}F^\times$, the exterior algebra on $F^\times$ sitting in degree one.

Now Matsumoto's theorem says that $K_2(F)$ is the quotient of $F^\times \otimes_{\mathbb{Z}}F^\times$ by the subgroup generated by elements of the form $a \otimes (1-a)$. These relations imply that $u\otimes v$ is equivalent to $v\otimes u$, so $K_2(F)$ is really a quotient of $\Lambda^2_{\mathbb{Z}}F^\times$. So there is a surjective symmetrising map from $\Lambda^2_{\mathbb{Z}} F^\times$ to $K_2(F)$ sending $u\wedge v$ to the Steinberg symbol $\{u, v\}$ represented by $u\otimes v$. Looking at [DS (A2)], this is the map in the Bloch-Wigner theorem. In particular, it comes from the inclusion of the split torus into $SL(2,F)$, see [DS, top of p189], so this is your map.

The conclusion is that your map surjects onto $K_2(F)$, but that $H_2(PGL(2,F),\mathbb{Z})$ contains an extra factor of two. Since $\Lambda^2_{\mathbb{Z}}F^\times$ is a $\mathbb{Q}$-vector space, I don't think the extra factor of two can be hit by $H_2(G,\mathbb{Z})$, but I admit that this confuses me.

This answer has been edited so that it not only reads better but also gives a better answer. I'll refer to Dupont and Sah, "Scissors congruences, II" as [DS] and the paper of Sah and Wagoner, "Second cohomology of Lie groups made discrete" as [SW] in what follows.

Let $F$ be an algebraically closed field of characteristic zero; the question asks about $F=\mathbb{C}$ but the answer is true in at least this generality. Then $PGL(2,F)\cong PSL(2,F)$ sits in an exact sequence $$1 \to \mathbb{Z}/2 \to SL(2,F) \to PGL(2,F) \to 1.$$ So $K_2(F)=H_2(SL(2,F),\mathbb{Z})$ and $H_2(PGL(2,F),\mathbb{Z})\cong K_2(F)\oplus \mathbb{Z}/2$ see [DS (A5)] and [SW].

Let $G$ be the upper triangular matrices in $PGL(2,F)$ and let $U$ be the subgroup of $G$ consisting of upper triangular matrices with ones on the diagonal. So there is a split exact sequence $1 \to U \to G \to F^\times \to 1$, and $U$ is isomorphic to the additive group of $F$.

It is shown in [DS (A11)] that the inclusion $F^\times \to G$ and the quotient map $G \to F^\times$ induce an isomorphism $H_*(F^\times,\mathbb{Z}) \cong H_*(G,\mathbb{Z})$. In particular, it follows that $H_1(F^\times,U)=0$ (in your case, this is $H_1(\mathbb{C}^\times,\mathbb{C})=0$). Thus $H_*(G,\mathbb{Z})\cong \Lambda^*_{\mathbb{Z}}F^\times$, the exterior algebra on $F^\times$ sitting in degree one.

Now Matsumoto's theorem says that $K_2(F)$ is the quotient of $F^\times \otimes_{\mathbb{Z}}F^\times$ by the subgroup generated by elements of the form $a \otimes (1-a)$. These relations imply that $u\otimes v$ is equivalent to $-v\otimes u$, so $K_2(F)$ is really a quotient of $\Lambda^2_{\mathbb{Z}}F^\times$. So there is a surjective symmetrising map from $\Lambda^2_{\mathbb{Z}} F^\times$ to $K_2(F)$ sending $u\wedge v$ to the Steinberg symbol $\{u, v\}$ represented by $u\otimes v$. Looking at [DS (A2)], this is the map in the Bloch-Wigner theorem. In particular, it comes from the inclusion of the split torus into $SL(2,F)$, see [DS, top of p189], so this is your map.

The conclusion is that your map surjects onto $K_2(F)$, but that $H_2(PGL(2,F),\mathbb{Z})$ contains an extra factor of two. Since $\Lambda^2_{\mathbb{Z}}F^\times$ is a $\mathbb{Q}$-vector space, I don't think the extra factor of two can be hit by $H_2(G,\mathbb{Z})$, but I admit that this confuses me.