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You have to take the homotopy colimit, which is the Grothendieck construction localized at a certain class of morphisms (arguably this is more computable than the 1-categorical colimit, so this is actually nice). It agrees up to equivalence with the ordinary 1-categorical colimit when the diagram is projectively cofibrant in the folk model structure on Cat.

So the answer is negative unless you take the appropriate 2-categorical universal construction as your definition of colimit.

However, an interesting thing to note is that if you work ∞-categorically, the (∞,1)-category of 1-categories (which is a (2,1)-category) does produce the correct answer (since in that case, all colimits are ∞-colimits). That's also why the model category picture produces the same answer.

You can also use the Bousfield-Kan formula as another presentation of the homotopy colimit, fwiw (see here).

Edit: Here's a silly counterexample:

Consider the diagram

$$\ast \leftarrow \ast \coprod \ast \to \ast$$

where $\ast$ denotes the terminal category. If we form the 1-categorical colimit, we clearly obtain a point.

However, this diagram is naturally equivalent to

$$J\hookleftarrow \ast\coprod\ast \hookrightarrow J$$

Where $J$ denotes the free-walking isomorphism (unique contractible groupoid with two objects). (The transformation here is the identity on the middle term and the terminal map on the outer terms. This is strictly natural, although its quasi-inverse is only pseudonatural).

In this case, the strict colimit is $$\Sigma (\ast \coprod \ast)=\operatorname{B}\mathbb{Z},$$ which is the correct homotopy colimit.

You have to take the homotopy colimit, which is the Grothendieck construction localized at a certain class of morphisms (arguably this is more computable than the 1-categorical colimit, so this is actually nice). It agrees up to equivalence with the ordinary 1-categorical colimit when the diagram is projectively cofibrant in the folk model structure on Cat.

So the answer is negative unless you take the appropriate 2-categorical universal construction as your definition of colimit.

However, an interesting thing to note is that if you work ∞-categorically, the (∞,1)-category of 1-categories (which is a (2,1)-category) does produce the correct answer (since in that case, all colimits are ∞-colimits). That's also why the model category picture produces the same answer.

You can also use the Bousfield-Kan formula as another presentation of the homotopy colimit, fwiw (see here).

Edit: Here's a silly counterexample:

Consider the diagram

$$\ast \leftarrow \ast \coprod \ast \to \ast$$

where $\ast$ denotes the terminal category. If we form the 1-categorical colimit, we clearly obtain a point.

However, this diagram is naturally equivalent to

$$J\hookleftarrow \ast\coprod\ast \hookrightarrow J$$

Where $J$ denotes the free-walking isomorphism (unique contractible groupoid with two objects). (The transformation here is the identity on the middle term and the terminal map on the outer terms. This is strictly natural, although its quasi-inverse is only pseudonatural).

In this case, the strict colimit is $$\Sigma (\ast \coprod \ast)=\operatorname{B}\mathbb{Z},$$ which is the correct homotopy colimit.

We can also verify that we get the same answer by doing the Grothendieck construction and localizing at the Cartesian arrows. If we call this diagram $D$, then the Grothendieck construction $$\int D$$ is the category

$$\begin{matrix} &&\ast&&\\ &\swarrow&&\searrow&\\ \ast && && \ast\\ &\nwarrow&&\nearrow&\\ &&\ast&& \end{matrix}$$

Localizing at the Cartesian morphisms (which are in this case all of the nonidentity morphisms), we obtain the associated totally localized groupoid, which is equivalent to $S^1=\operatorname{B}\mathbb{Z}$.

You have to take the homotopy colimit, which is the Grothendieck construction localized at a certain class of morphisms (arguably this is more computable than the 1-categorical colimit, so this is actually nice). It agrees up to equivalence with the ordinary 1-categorical colimit when the diagram is projectively cofibrant in the folk model structure on Cat.

So the answer is negative unless you take the appropriate 2-categorical universal construction as your definition of colimit.

However, an interesting thing to note is that if you work ∞-categorically, the (∞,1)-category of 1-categories (which is a (2,1)-category) does produce the correct answer (since in that case, all colimits are ∞-colimits). That's also why the model category picture produces the same answer.

You can also use the Bousfield-Kan formula as another presentation of the homotopy colimit, fwiw (see here).

Edit: Here's a silly counterexample:

Consider the diagram

$$\ast \leftarrow \ast \coprod \ast \to \ast$$

where $\ast$ denotes the terminal category. If we form the 1-categorical colimit, we clearly obtain a point.

However, this diagram is naturally equivalent to

$$J\leftarrow \ast\coprod\ast \to J$$

Where $J$ denotes the free-walking isomorphism. (The transformation here is the identity on the middle term and the terminal map on the outer terms. This is strictly natural, although its quasi-inverse is only pseudonatural).

In this case, the strict colimit is $$\Sigma (\ast \coprod \ast)=\operatorname{B}\mathbb{Z},$$ which is the correct homotopy colimit.

You have to take the homotopy colimit, which is the Grothendieck construction localized at a certain class of morphisms (arguably this is more computable than the 1-categorical colimit, so this is actually nice). It agrees up to equivalence with the ordinary 1-categorical colimit when the diagram is projectively cofibrant in the folk model structure on Cat.

So the answer is negative unless you take the appropriate 2-categorical universal construction as your definition of colimit.

However, an interesting thing to note is that if you work ∞-categorically, the (∞,1)-category of 1-categories (which is a (2,1)-category) does produce the correct answer (since in that case, all colimits are ∞-colimits). That's also why the model category picture produces the same answer.

You can also use the Bousfield-Kan formula as another presentation of the homotopy colimit, fwiw (see here).

Edit: Here's a silly counterexample:

Consider the diagram

$$\ast \leftarrow \ast \coprod \ast \to \ast$$

where $\ast$ denotes the terminal category. If we form the 1-categorical colimit, we clearly obtain a point.

However, this diagram is naturally equivalent to

$$J\hookleftarrow \ast\coprod\ast \hookrightarrow J$$

Where $J$ denotes the free-walking isomorphism (unique contractible groupoid with two objects). (The transformation here is the identity on the middle term and the terminal map on the outer terms. This is strictly natural, although its quasi-inverse is only pseudonatural).

In this case, the strict colimit is $$\Sigma (\ast \coprod \ast)=\operatorname{B}\mathbb{Z},$$ which is the correct homotopy colimit.

You have to take the homotopy colimit, which is the Grothendieck construction localized at a certain class of morphisms (arguably this is more computable than the 1-categorical colimit, so this is actually nice). It agrees up to equivalence with the ordinary 1-categorical colimit when the diagram is projectively cofibrant in the folk model structure on Cat.

So the answer is negative unless you take the appropriate 2-categorical universal construction as your definition of colimit.

However, an interesting thing to note is that if you work ∞-categorically, the (∞,1)-category of 1-categories (which is a (2,1)-category) does produce the correct answer (since in that case, all colimits are ∞-colimits). That's also why the model category picture produces the same answer.

You can also use the Bousfield-Kan formula as another presentation of the homotopy colimit, fwiw (see here).

You have to take the homotopy colimit, which is the Grothendieck construction localized at a certain class of morphisms (arguably this is more computable than the 1-categorical colimit, so this is actually nice). It agrees up to equivalence with the ordinary 1-categorical colimit when the diagram is projectively cofibrant in the folk model structure on Cat.

So the answer is negative unless you take the appropriate 2-categorical universal construction as your definition of colimit.

However, an interesting thing to note is that if you work ∞-categorically, the (∞,1)-category of 1-categories (which is a (2,1)-category) does produce the correct answer (since in that case, all colimits are ∞-colimits). That's also why the model category picture produces the same answer.

You can also use the Bousfield-Kan formula as another presentation of the homotopy colimit, fwiw (see here).

Edit: Here's a silly counterexample:

Consider the diagram

$$\ast \leftarrow \ast \coprod \ast \to \ast$$

where $\ast$ denotes the terminal category. If we form the 1-categorical colimit, we clearly obtain a point.

However, this diagram is naturally equivalent to

$$J\leftarrow \ast\coprod\ast \to J$$

Where $J$ denotes the free-walking isomorphism. (The transformation here is the identity on the middle term and the terminal map on the outer terms. This is strictly natural, although its quasi-inverse is only pseudonatural).

In this case, the strict colimit is $$\Sigma (\ast \coprod \ast)=\operatorname{B}\mathbb{Z},$$ which is the correct homotopy colimit.