2 two different places to forget to

Let $C$ be a category with a zero object, i.e. an object 0 which is both initial and terminal. Then $C$ is automatically (and uniquely) enriched over the category $Set_\star$ of pointed sets with smash product, where the basepoint $0\in C(x,y)$ is the unique map which factors through the object 0. Also, a functor between such categories is $Set_\star$-enriched (i.e. preserves the basepoints of homsets) if and only if it preserves the zero object. Call these pointed categories and pointed functors. (This is the "zero-ary" version of additive categories and biproducts.)

Now let $G:C^{op} \to Set_\star$ and $F:C\to Set_\star$ be pointed functors; we can then consider the tensor product $G \otimes_C F$ (otherwise known as the $G$-weighted colimit of $F$) either (1) as ordinary unenriched functors, or (2) as $Set_\star$-enriched functors. The first is a quotient of $\bigvee_c G(c)\times F(c)$, while the second is a quotient instead of $\bigvee_c G(c) \wedge F(c)$. (I'm writing $\vee$ for the coproduct in $Set_\star$, since it is a "wedge" which identifies basepoints in the disjoint union.) I believe, however, that these two tensor products turn out to be the same, since any pair $(x,0)$ in $G(c)\times F(c)$ is equal to $(x,F(t)(0))$ where $t:0\to c$ is the unique map from 0 to $c$ in $C$, and hence gets identified in the tensor product with $(G(t)(x),0) = (0,0)$, which is the basepoint of $\bigvee_c G(c)\times F(c)$. Thus when $C$ has a zero object, the ordinary tensor product over $C$ has the effect of automatically performing the smash product as well.

My question is: does this remain true for homotopy tensor products? Suppose we consider $Set_\star$ as sitting inside the category $sSet_\star$ of pointed simplicial sets, and instead of the tensor product we take its homotopical replacement, which can be described as a two-sided bar construction. We thus get two pointed simplicial sets $B^u(G,C,F)$ and $B^p(G,C,F)$ (for unpointed and pointed), defined by $$B^u_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \times C(c_{n-1},c_n)\times\dots\times C(c_0,c_1)\times F(c_0)$$ and $$B^p_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \wedge C(c_{n-1},c_n)\wedge\dots\wedge C(c_0,c_1)\wedge F(c_0)$$ There is a canonical quotient map $B^u(G,C,F) \to B^p(G,C,F)$, and the above observation (assuming it is correct) means that this map induces an isomorphism on $\pi_0$. Is it a weak homotopy equivalence?

Edit: Reading over the answers, I realized that there are actually two different questions here. Greg answered a question which isn't quite what I asked, but fortunately the question he answered is the one that I meant to ask, which is comparing $B^p(G,C,F)$ with the homotopy tensor product when considering $G$ and $F$ as functors landing in unpointed simplicial sets, so that the $\bigvee$s would actually become disjoint unions $\coprod$.

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# Pointed vs. unpointed homotopy colimits

Let $C$ be a category with a zero object, i.e. an object 0 which is both initial and terminal. Then $C$ is automatically (and uniquely) enriched over the category $Set_\star$ of pointed sets with smash product, where the basepoint $0\in C(x,y)$ is the unique map which factors through the object 0. Also, a functor between such categories is $Set_\star$-enriched (i.e. preserves the basepoints of homsets) if and only if it preserves the zero object. Call these pointed categories and pointed functors. (This is the "zero-ary" version of additive categories and biproducts.)

Now let $G:C^{op} \to Set_\star$ and $F:C\to Set_\star$ be pointed functors; we can then consider the tensor product $G \otimes_C F$ (otherwise known as the $G$-weighted colimit of $F$) either (1) as ordinary unenriched functors, or (2) as $Set_\star$-enriched functors. The first is a quotient of $\bigvee_c G(c)\times F(c)$, while the second is a quotient instead of $\bigvee_c G(c) \wedge F(c)$. (I'm writing $\vee$ for the coproduct in $Set_\star$, since it is a "wedge" which identifies basepoints in the disjoint union.) I believe, however, that these two tensor products turn out to be the same, since any pair $(x,0)$ in $G(c)\times F(c)$ is equal to $(x,F(t)(0))$ where $t:0\to c$ is the unique map from 0 to $c$ in $C$, and hence gets identified in the tensor product with $(G(t)(x),0) = (0,0)$, which is the basepoint of $\bigvee_c G(c)\times F(c)$. Thus when $C$ has a zero object, the ordinary tensor product over $C$ has the effect of automatically performing the smash product as well.

My question is: does this remain true for homotopy tensor products? Suppose we consider $Set_\star$ as sitting inside the category $sSet_\star$ of pointed simplicial sets, and instead of the tensor product we take its homotopical replacement, which can be described as a two-sided bar construction. We thus get two pointed simplicial sets $B^u(G,C,F)$ and $B^p(G,C,F)$ (for unpointed and pointed), defined by $$B^u_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \times C(c_{n-1},c_n)\times\dots\times C(c_0,c_1)\times F(c_0)$$ and $$B^p_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \wedge C(c_{n-1},c_n)\wedge\dots\wedge C(c_0,c_1)\wedge F(c_0)$$ There is a canonical quotient map $B^u(G,C,F) \to B^p(G,C,F)$, and the above observation (assuming it is correct) means that this map induces an isomorphism on $\pi_0$. Is it a weak homotopy equivalence?